1

6 TRIANGLES Exercise 6.1

Q.1. Fill in the blanks using the correct word given in brackets: (i) All circles are ______. (congruent, similar) (ii) All squares are ______. (similar, congruent) (iii) All ______ triangles are similar. (isosceles, equilateral) (iv) Two polygons of the same number of sides are similar, if (a) their corresponding angles are ______ and (b) their corresponding

sides are ______. (equal, proportional)

Solution. (i) All circles are similar. (ii) All squares are similar. (iii) All equilateral triangles are similar. (iv) Two polygons of the same number of sides are similar, if (a) their corresponding angles are equal and (b) their corresponding sides are proportional.

Q.2. Give two different examples of pair of (i) similar figures. (ii) non-similar figures.

Solution. (i) Two different examples of similar figures are: (a) Two twenty-rupees notes.

(b) Two two-rupees coins. (ii) Two different examples of non-similar figures are: (a) One-rupee coin and a five rupees coin. (b) One-rupee note and ten rupees note.

Q.3. State whether the following quadrilaterals are similar or not:

2

Solution. The two quadrilaterals in the given figure are not similar because their corresponding angles are not equal.

Exercise 6.2 Q.1. In figure (i) and (ii), DE || BC. Find EC in (i) and AD in (ii).

Solution. (i) in BC||DEABC,Δ (Given)

Therefore, ECAE

DBAD

= [By using Basic Proportionality Theorem]

⇒ EC1

35.1

=

⇒ Σ5.1

3EC =

215

103EC =×

= cm

Hence, cm 2EC = (ii) In ABC,Δ

BC||DE (Given)

Therefore, ECAE

DBAD

= [By using Basic Proportionality Theorem]

⇒ 4.58.1

7.2AD

=

⇒ 1.8 7.2 18 72 10 24AD5.4 10 10 54 10×

= = × × =

⇒ 4.2AD = Hence, cm. 4.2AD =

3

Q.2. E and F are points on the sides PQ and PR respectively of a ΔPQR . For each of the following cases, state whether EF || OR : (i) PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2.4 cm (ii) PE = 4 cm, QE = 4.5 cm, PF = 8 cm and RF = 9 cm (iii) PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.36 cm

Solution. In PQRΔ , E and F are two points on side PQ and PR respectively (i) cm 3EQ cm, 9.3PE == (Given)

cm 4.2FR cm, 6.3PF == (Given)

Therefore, 3.11013

3039

39.3

EQPE

====

And, 5.123

2436

4.26.3

FRPF

====

So, FRPF

EQPE

≠

Hence, EF is not parallel to QR. (ii) PE 4 cm, QE 4.5 cm, PF 8 cm, RF 9 cm= = = =

Therefore, 98

4540

5.44

QEPE

=== (By using Basic Proportionality Theorem)

And, PF 8RF 9

=

So, RFPF

QEPE

=

Hence, EF is parallel to QR (iii) cm 0.36PF cm, 0.18PE cm, 2.56PR cm, 1.28PQ ==== (Given)

Here, cm 1.100.181.28PEPQEQ =−=−= And, FR PR PF 2.56 0.36 2.20 cm= − = − =

(By using Basic Proportionality Theorem)

4

So, PE 0.18 18 9EQ 1.10 110 55

= = = ...(i)

And, 559

22036

20.236.0

FRPE

=== ...(ii)

Therefore, PE PFEQ FR

=

Hence, EF is parallel to QR.

Q.3. In figure, LM || CB and LN || CD, prove that

AM AN= .AB AD

Solution. In ABC,Δ we have BC||ML (Given)

⇒ LCAL

MBAM

= (By using Basic Proportionality theorem) ...(i)

In ACD,Δ we have DC||LN (Given)

⇒ AN ALND LC

= (By using Basic Proportionality Theorem) ...(ii)

From equation (i) and equation (ii), we get

NDAN

MBAM

=

or, ANND

AMMB

=

or, 1ANND1

AMMB

+=+ (Adding 1 in both sides)

or, AN

ANNDAM

AMMB +=

+

or, ANAD

AMAB

=

So, ADAN

ABAM

=

5

Q.4. In figure DE || AC, and DF || AE, prove that BF BE= .FE EC

Solution. In AC||DE ABC,Δ (Given)

Therefore, ECBE

DABD

= ...(i) (By using Basic Proportionality Theorem)

In ABE,Δ AE||DF (Given)

Therefore, FEBF

DABD

= ...(ii) (By using Basic Proportionality Theorem)

From equation (i) and equation (ii), we have

FEBF

ECBE

=

Q.5. In figure DE || OQ and DF || OR. Show that EF || QR.

Solution. Given: DE || OQ and DF || OR To prove: s QR.||EF Proof: In PQO,Δ DE || OQ (Given)

Therefore, EQPE

DOPD

= ...(i) (By using Basic Proportionality Theorem)

In POR,Δ DF || OR

Therefore, FRPF

DOPD

= ...(ii) (By using Basic Proportionality Theorem)

From equation (i) and equation (ii), we have

FRPF

EQPE

=

In PQR,Δ by using converse of Basic Proportionality Theorem. EF is parallel to QR.

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Q.6. In figure, A, B and C are points on OP, OQ and OR respectively such that AB || PQ and AC || PR. Show that BC || QR.

Solution. Given: PQR,Δ A, B and C are points on OP, OQ and OR respectively such that

PR||AC and PQ||AB To prove: QR.||BC Proof: In OPQ,Δ we have PQ||AB (Given)

Therefore, BQOB

APOA

= ...(i) (By using Basic Proportionality Theorem)

In OPR,Δ we have PR||AC (Given)

Therefore, CROC

APOA

= ...(ii) (By using Basic Proportionality Theorem)

From equation (i) and (ii), we have

CROC

BQOB

=

Hence, in OQR, BC||QR,Δ (By Converse of Basic Proportionality Theorem)

Q.7. Using Basic Proportionality Theorem, prove that a line drawn through the mid-point of one side of a triangle parallel to another side bisects the third side. (Recall that you have proved it in class IX).

Solution. Given: D ABC,Δ is the mid point of AB, i.e., DB.AD = A line parallel to BC intersects AC at E as shown in figure, i.e.,

BC.||DE To prove: E is the mid point of AC. Proof: D is the mid point of AB. Therefore, DBAD =

⇒ 1BDAD

= ...(i)

In BC||DE ABC,Δ

Therefore, ECAE

DBAD

= (By Basic Proportionality Theorem)

7

⇒ ECAE1 = [From equation (i)]

Therefore, AE EC= Hence, E is the mid-point of AC, proved.

Q.8. Using converse of Basic Proportionality Theorem, prove that the line joining the mid-points of any two sides of a triangle is parallel to the third side. (Recall that you have done it in class IX).

Solution. Given: E and D ABC,Δ are the mid points of AB and AC respectively such that BDAD = and EC,AE =

To prove: BC||DE Proof: D is the mid-point of AB (Given) Therefore, BDAD =

⇒ 1BDAD

= ...(i)

Aslo, E is the mid-point of AC (Given) Therefore, ECAE =

1ECAE

= ...(ii)

From equation (i) and equation (ii), we have ECAE

BDAD

=

Hence, BC||DE , (By converse of Basic Proportionality Theorem)

Q.9. ABCD is a trapezium in which AB || DC and its diagonals intersect each other

at the point O. Show that AO CO= .BO DO

Solution. Given: ABCD is a trapezium in which DC,||AB diagonals AC and BD intersect each other at O.

To prove: AO COBO DO

=

Construction: Through O, draw AB||DC||EO Proof: In ADC,Δ we have DC||EO (By construction)

Therefore, DEAE

COAO

= ...(i) (By using Basic Proportionality Theorem)

In DBAΔ , we have AB||EO (By construction)

8

BODO

EADE

= ...(ii) (By using Basic Proportionality Theorem)

From equation (i) and equation (ii), we get

AO BOCO DO

=

or, DOCO

OBAO

=

Q.10. The diagonals of a quadrilateral ABCD intersect each other at the point O

such that AO CO= .BO DO

Show that ABCD is a trapezium.

Solution. Given: Quadrilateral ABCD, diagonals AC and BD intersects each other at O

such that .DOCO

BOAO

=

To Prove: ABCD is trapezium.

Construction: Through O, draw line EO, where AB||EO which meets AD at E.

Proof: In DAB,Δ we have

AB||EO

Therefore, OBDO

EADE

= ...(i) (By using Basic Proportionality Theorem)

Also, DOCO

BOAO

= (Given)

or, DOBO

COAO

=

or, BODO

AOCO

=

⇒ AOCO

OBDO

= ...(ii)

From equation (i) and equation (ii), we get

AOCO

EADE

=

Therefore, By using converse of Basic Proportionality Theorem, DC||EO also EO||AB ⇒ DC.||AB

Hence, quadrilateral ABCD is a trapezium with AB || CD.

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Exercise 6.3 Q.1. State which pairs of triangles in Figure are similar. Write the similarity

criterion used by you for answering the question and also write the pairs of similar triangles in the symbolic form:

10

Solution. (i) In PQR, and ABC ΔΔ We have °=∠=∠ 60PA (Given) °=∠=∠ 80QB (Given) °=∠=∠ 40RC (Given)

Therefore, PQR~ABC ΔΔ (AAA similarity criterion)

In ΔABC and ΔPQR, We have (ii) AB 2, BC 2.5, AC 3, QR 4, PR 5, PQ 6= = = = = =

21

42

RQAB

== ...(i)

21

63

PQAC

== ...(ii)

21

55.2

PRBC

== ...(iii)

From equation (i), equation (ii) and equation (iii), we have

21

PRBC

PQAC

RQAB

===

Therefore, QRP~ABC ΔΔ (SSS similarity criterion) (iii) In LMPΔ and DEFΔ , We have

6DF 4,DE 5,EF 3,LP 2,MP 2.7,LM ======

21

42

DEMP

==

11

21

63

DFPL

==

5027

57.2

EFLM

==

Here, EFLM

DFPL

DEMP

≠= .

Hence, DEF and LMP ΔΔ are not similar. (iv) In PQR, and MNL ΔΔ We have

MN 3, ML 5, QR 10, PQ 5, M 70 , A 70= = = = ∠ = ° ∠ = °

Here, 21

105

QRML

==

Aslo, °=∠=∠ 70QM

And, 21

63

PQMN

==

Hence, QPR~MNL ΔΔ (By SAS similarity criterion) (v) In DEF, and ABC ΔΔ We have

AB 2.5, BC 3, A 80 , EF 6, DF 5, F 80= = ∠ = ° = = ∠ = °

Here, 21

55.2

DFAB

==

And, 21

63

EFBC

==

⇒ FB ∠≠∠ Hence, DEF and ABC ΔΔ are not similar.

(vi) In DEFΔ , We have °=∠+∠+∠ 180FED (sum of angles of a triangle) ⇒ °=∠+°+° 180F8070

°=∠⇒°−°−°=∠⇒

30F 8070180F

In PQR,Δ We have 180RQP =∠+∠+∠ (Sum of ) of Angles Δ ⇒ °=°+°+∠ 1803080P ⇒ °−°−°=∠ 3080180P ⇒ °=∠ 70P In PQR, and DEF ΔΔ We have °=∠=∠ 70PD

12

°=∠=∠ 80QE °=∠=∠ 30RF Hence, PQR~DEF ΔΔ (By AAA similarity criterion)

Q.2. In Figure, ∠ΔODC ~ ΔOBA, BOC = 125° and ∠CDO = 70°. Find DCO DOC, ∠∠ and ∠OAB.

Solution. Given that: °=∠ 125BOC °=∠ 70CDO

Therefore, DOC COB 180∠ + ∠ = ° (Linear pair axiom) ⇒ DOC 125 180∠ + ° = °

DOC 180 125 DOC 55

⇒ ∠ = ° − °⇒ ∠ = °

⇒ ∠DOC = ∠AOB = 55° (Vertically opposite angles) But OBA~ODC ΔΔ (Given)

Therefore, D B 70∠ = ∠ = ° In DOC, D DOC C 180Δ ∠ + ∠ + ∠ = °

⇒ 70 55 C 180° + ° + ∠ = ° ⇒ C 180 70 55∠ = ° − ° − ° ⇒ C 55∠ = ° C A 55∠ = ∠ = ° [Since, ΔODC ~ ΔOBA] Hence, 55DOC =∠ ° 55DCA =∠ ° 55OAB =∠ °

Q.3. Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at the point O. Using a similarity criterion for two triangles, show that OA OB= .OC OD

Solution. Given: Trapezium ABCD, CD||AB and diagonals AC and BD intersects each other at O.

To prove: ODOB

OCAO

=

Proof: In OCD, and OAB ΔΔ We have CD||AB Here, OCDOAB ∠=∠ (alternate angles) CODAOB ∠=∠ (vertically opposite angle) And, ODCOBA ∠=∠ (alternate angles)

13

Therefore, OCD~OAB ΔΔ (AAA Similarity, Criteion)

⇒ OAOC

BODO

=

Hence, DOBO

OCOA

=

Q.4. In figure PRQT

QSQR

= and 2.1 ∠=∠ Show that ΔTQR.~ΔPQS

Solution. Given that,

PRQT

QSQR

= and 21 ∠=∠

To prove: TQR~PQS ΔΔ Proof: In PQR,Δ We have 21 ∠=∠ (Given) Therefore, PQPR = (sides opposite to equal angles of a Δ are equal) ...(i)

and, PRQT

QSQR

= (Given)

or, QPQT

QSQR

= PQ)(PR = ...(ii)

⇒ QTPQ

QRQS

=

In TQR, and PQS ΔΔ We havess

QTPQ

QRQS

=

Also, RQTSQP ∠=∠ ⇒ 11 ∠=∠ Hence, TQR~PQS ΔΔ (SSS similarity criterion)

Q.5. S and T are points on sides PR and QR of PQRΔ such that ∠ ∠P = RTS. Show that ΔRPQ ~ ΔRTS.

Solution. In T and S PQR,Δ are points on side PR and QR of PQRΔ such that RTSP ∠=∠

To prove:S RTS~RPQ ΔΔ

Proof: In RTS and RPQ ΔΔ

RTSRPQ ∠=∠ (Given)

[If two triangles are similar then corresponding sides are proportional)

14

RR ∠=∠ s (Common angle)

Therefore, RTS~RPQ ΔΔ (By AA similarity criterion)

Q.6. In figure if ΔABE ΔACD,≅ show that ΔADE ~ ΔABC.

Solution. Given: ABCΔ in which ACDABE Δ≅Δ To prove: ABC~ADE ΔΔ Proof: ACDABE Δ≅Δ (Given) ⇒ ACAB = (by cpct) And, ADAE = (by cpct)

⇒ 1ACAB

= ...(i)

1ADAE

= ...(ii)

From equation (i) and (ii), We have

AEAD

ACAB

=

In ABC, and ADE ΔΔ We have

ACAB

AEAD

=

And, AA ∠=∠ (common) Hence, ABC~ADE ΔΔ (By SAS similarity criterion)

Q.7. In the given figure, altitudes AD and CE of ΔABC intersect each other at the point P. Show that: (i) ΔAEP ~ ΔCDP (ii) ΔABD ~ ΔCBE (iii) ΔAEP ~ ΔADB (iv) ΔPDC ~ ΔBEC

Solution. Given: AB,CE BC,AD ABC, ⊥⊥Δ To prove: (i) CDP~AEP ΔΔ (ii) CBE~ABD ΔΔ (iii) ADB~AEP ΔΔ (iv) BEC~PDC ΔΔ Proof: (i) In CDP, and AEP ΔΔ We have AEP CDP 90∠ = ∠ = ° ...(i) And, EPA DPC∠ = ∠ ...(ii) (Vertically opposite angle)

15

From equation (i) and equation (ii), we have ∴ CDP~AEP ΔΔ (By AA Similarity Criterion)

(ii) In CBE, and ABD ΔΔ 90CEBADB =∠=∠ ° ...(i) And, CBEABD ∠=∠ ...(ii) From equation (i) and equation (ii), We have CBE~ABD ΔΔ (By AA similarity criterion)

(iii) In ADB, and AEP ΔΔ ADBAEP ∠=∠ = 90° ...(i) DABEAP ∠=∠ ...(ii) From equation (i) and equation (ii), we have

ADB~AEP ΔΔ (By AA similarity criterion) (vi) In BEC, and PDC ΔΔ We have ECBDCP ∠=∠ ...(i) (common) PDC BEC 90∠ = ∠ = ° ...(ii) Hence, BEC~PDC ΔΔ (By AA similarity criterion)

Q.8. E is a point on the side AD produced of a parallelo-gram ABCD and BE intersects CD at F. Show that ΔABE ~ ΔCFB.

Solution. Given: Parallelogram ABCD. Side AD is produced to E and BE intersects DC at F.

To Prove: CFB~ABE ΔΔ Proof: In CFB. and ABE ΔΔ , We have FCBBAF ∠=∠ ...(i) (opposite angles of || gm) CBFAEB ∠=∠ ...(ii) (alternate angles) From equation (i) and equation (ii), we have CFB~ABE ΔΔ (By AA similarity criterion)

Q.9. In the given figure, ABC and AMP are two right triangles, right angled at B and M respectively. Prove that: (i) ΔABC ~ ΔAMP

(ii) CA BC= .PA MP

Solution. Given: ABCΔ and AMPΔ are two right angled triangles at B and M respectively. To Prove: (i) AMP~ABC ΔΔ

(ii) MPBC

PACA

=

16

Proof: (i) In ABCΔ and AMP,Δ We havess AA ∠=∠ (common angle) ∠ABC = ∠AMP = 90° (each 90°) Therefore, ΔABC ~ ΔAMP (By AA similarity criterion) (ii) As, ∠ABC = ∠ΣAMP If two triangle are similar, then corresponding sides are equal

Hence, MPBC

PACA

=

Q.10. CD and GH are respectively the bisectors of ∠ACB and ∠EGF such that D and H lie on sides AB and FE of ΔABC and ΔEFG respectively. If ΔABC ~ ΔFEG, show that:

(i) CD AC=GH FG

ss

(ii) ΔDCB ~ ΔHGE (iii) ΔDCA ~ ΔHGF

Solution. Given: In ABCΔ and EFG,Δ CD and GH are bisectors of ∠ACB and ∠EGF respectively

21 ∠=∠ And, 43 ∠=∠ FEG~ABC ΔΔ

To prove: (i) FGAC

GHCD

=

(ii) HGE~DCB ΔΔ (iii) HGF~DCA ΔΔ Proof: (i) Given that, FEG~ABC ΔΔ

Therefore, EB F;A ∠=∠∠=∠ and GC ∠=∠ [∵ The corresponding angles are equal] Consider, GC ∠=∠ [Proved above]

⇒ G21C

21

∠=∠

⇒ 42 ∠=∠ or 31 ∠=∠ Now, in ACD and FGHΔ Δ , We have FA ∠=∠ [Proved above]

17

⇒ 42 ∠=∠ [Proved above] Therefore, FGH~ACD ΔΔ [By AA similarity criterion]

Hence, FGAG

GHCD

= [Since, corresponding sides are proportional]

(ii) In HGE, and DCB ΔΔ We have EB ∠=∠ [Proved above] And, 31 ∠=∠ [Proved above] Hence, HGE~DCB ΔΔ [By AA similarity criterion]

(iii) In DCA and HGF,Δ Δ We have A F and 2 4∠ = ∠ ∠ = ∠ Hence, DCA~ HGFΔ Δ [By AA similarity criterion]

Q.11. In the given Figure, E is a point on side CB produced of an isosceles triangle ABC with AB = AC, If AD BC⊥ and EF AC,⊥ prove that ΔABD ~ ΔECF.

Solution. Given: An isosceles triangle ABC with BC,AC AC,AB ⊥= side BC is produced to E. ACEF ⊥

To prove: ECF~ABD ΔΔ

Proof: [ABC is an isosceles triangle 6]

ACAB =

Therefore, ∠ABC = ∠ACB [Equal sides have equal angles opposite to it]

Or, ∠ABD = ∠ECF

In ECF, and ABD ΔΔ

ECFABD ∠=∠ (Proved above)

ADB EFC 90∠ = ∠ = °

Hence, ECF~ABD ΔΔ [By AA similarity criterion]

Q.12. Sides AB and BC and median AD of a triangle ABC are respectively proportional to sides PQ and QR and median PM of ΔPQR (see figure). Show that ΔABC ~ ΔPQR.

Solution. Given: PQR, and ABC ΔΔ AB, BC, and median AD of ABCΔ are proportional to sides PQ, QR and median PM of PQR,Δ

i.e., PMAD

QRBC

PQAB

==

To Prove: PQR~ABC ΔΔ

18

Proof: AB BC ADPQ QR PM

= =

⇒ PMAD

QR21

BC21

PQAB

== ...(i)

⇒ PMAD

QMBD

PQAB

== (D is the mid-point of BC. M is the mid-point of QR)

⇒ PQM~ABD ΔΔ [SSS similarity criterion]

Therefore, PQMABD ∠=∠

⇒ PQR,ABC ∠=∠

In PQR and ABC ΔΔ

QRBC

PQAB

= ...(i)

PQRABC ∠=∠ ...(ii)

From equation (i) and equation (ii), we have PQR.~ABC ΔΔ [By SAS similarity criterion]

Q.13. D is a point on the side BC of a triangle ABC such that ∠ ∠ADC = BAC. Show that ⋅2CA = CB CD.

Solution. Given: ABC,Δ D is a point on side BC such that BAC.ADC ∠=∠

To Prove: CDBCCA2 ×= Proof: In ABCΔ and ADC.Δ CC ∠=∠ (Common) ADCBAC ∠=∠ (Given)

Therefore, DAC~ABC ΔΔ [By AA similarity criterion]

Therefore, ACBC

DCAC

=

Hence, CDCBCA2 ⋅=

Q.14. Sides AB and AC and median AD of a triangle ABC are respectively proportional to sides PQ and PR and median PM of another triangle PQR. Show that ΔABC ~ ΔPQR.

[Corresponding angles of two similar triangles are equal]

[If two triangles are similar corresponding sides are proportional]

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Solution. Given: Two triangles ABC and PQR in which AD and PM are medians such

that ABPQ

= ACPR

= ADPM

To prove : ΔABC ~ ΔDEF

Construction : Produce AD to E so that AD = DE. Join CE, Similarly produce PM to N such that PM = MN, also join RN.

Proof : In ΔABD and ΔCDE, We have AD = DE [By construdtion] BD = DC [∴ AP is the median] and, ∠ADB = ∠CDE [Vertically opp. angles] Therefore, ΔABD ≅ ΔCDE [By SAS Criterion of Congruence] ⇒ AB = CE [CPCT] …(i) Aslo, in ΔPQM and ΔMNR, We have PM = MN [By Construction] QM = MR [∴ PM is the median] and, ∠PMQ = ∠NMR [Vertically opp. ∠S] Therefore, ΔPQM = ΔMNR [By SAS Criterion of congruence] ⇒ PQ = RN [CPCT] …(ii)

Now, ABPQ

= ACPR

= ADPM

⇒ CERN

= ACPR

= ADPM

….[From (i) and (ii)]

⇒ CERN

= ACPR

= 2AD2PM

⇒ CERN

= ACPR

= AEPN

[∴ 2AD = AE and 2PM = PN]

∴ ΔACE ~ ΔPRN [By SSS Similarly Condition] Therefore, ∠2 = ∠4 Similarly, ∠1 = ∠3 Therefore, ∠ 1 + ∠ 2 = ∠3 + ∠4

20

⇒ ∠A = ∠P …….. (iii) Now, In ΔABC and ΔPQR, We have

ABPQ

= ACPR

[Given]

∠A = ∠P [From (iii)] Therefore, ΔABC ~ ΔPQR [By SAS Similarly Condition]

Q.15. A vertical pole of length 6 m casts a shadow 4 m long on the ground and at the same time a tower casts a shadow 28 m long. Find the height of the tower.

Solution. Length of the vertical pole 6= m (Given) Shadow of the pole 4= m (Given)

Let height of tower h= m Length of shadow of tower 28= m (Given) In ABC and DFEΔ Δ EC ∠=∠ (angular elevation of sum) 90FB =∠=∠ Therefore, ABC~ DEFΔ Δ (By AA similarity criterion)

Therefore, EFBC

DFAB

=

Therefore, 2846

=h

⇒ 4286×

=h

⇒ 76×=h Or, 42=h m Hence, height of the tower 42 m.=

Q.16. If AD and PM are medians of triangles ABC and PQR, respectively where

ΔABC ~ ΔPQR, prove that AB AD= .PQ PM

Solution. Given: AD and PM are median the medians of ΔABC and ΔPQR respectively and ABC~ PQRΔ Δ

To Prove: AB ADPQ PM

=

Proof: PQR~ABC ΔΔ (Given)

Therefore, PRAC

QRBC

PQAB

==

PA ∠=∠

[If two triangles are similar corresponding sides are proportional]

[If two triangles are similar corresponding sides are proportional]

21

QB ∠=∠ and, RC ∠=∠ As D is the mid-point of BC

So, BC21DCBD == ...(i)

Also, M is the mid-point of QR.

So, QR21MRQM == ...(ii)

⇒ QRBC

PQAB

= [∴ ΔABC ~ ΔPQR]

⇒ 2QM2BD

PQAB

= [From equation (i) and equation (ii)]

⇒ QMBD

PQAB

=

And, PQMABD ∠=∠ (Given) Therefore, PQM~ABC ΔΔ (By SAS similarity criterion)

Hence, PMAD

PQAB

=

Exercise 6.4

Q.1. Let ΔABC ~ ΔDEF and their areas be, respectively, 64 2cm and 121 2cm . If EF = 15.4 cm, find BC.

Solution. It is given that, Area of 2cm 64ABC =Δ Area of 2cm 121DEF =Δ cm 15.4EF = And, DEF~ABC ΔΔ

Therefore, 2

2Area of ABC ABArea of DEF DE

Δ=

Δ

2 2

2 2AC BCDF EF

= = ...(i)

[If two triangles are similar, ratio of their areas are equal to square of the ratio of their corresponding sides]

[If two triangles are similar corresponding sides are proportional]

22

Therefore, 2

2

EFBC

12164

=

⇒ 22

15.4BC

118

⎟⎠⎞

⎜⎝⎛=⎟

⎠⎞

⎜⎝⎛

⇒ ⇒ 15.4BC

118

=

⇒ 1115.48BC ×

=

⇒ 1.48BC ×= ⇒ 11.2BC = cm

Q.2. Diagonals of a trapezium ABCD with AB || DC interesect each other at the point O. If AB = 2 CD, find the ratio of the areas of triangles AOB and COD.

Solution. ABCD is a trapezium with DC.||AB Diagonals AC and BD intersect each other at the point O.

In COD, and AOB ΔΔ We have 21 ∠=∠ (Alternate angles) 43 ∠=∠ (Alternate angles) 65 ∠=∠ (Vertically opposite angle) Therefore, COD~AOB ΔΔ [By AAA similarity criterion]

Now, 2

2

CDAB

COD)( of AreaAOB)( of Area

=ΔΔ

2

2

CDCD)2(

= [∴ AB = 2CD]

Therefore, 14

CDCD4

COD)( of AreaAOB)( of Area

2

2

==ΔΔ

Hence, required ratio of area of AOBΔ and 1 : 4COD =Δ

Q.3. In the figure, ABC and DBC are two triangles on the same base BC. If AD intersects BC at O, show that

ar (ABC) AO= .ar (DBC) DO

Solution. Given: ABC and DBC are the triangles on the same base BC. AD intersects BC at O.

[If two triangles are similar then ratio oftheir areas are equal to the square of theratio of their corresponding sides]

23

To Prove: DOAO

DBC)( of AreaABC)( of Area

=ΔΔ

Construction: Draw BCDM BC,AL ⊥⊥ Proof: In DMO, and ALO ΔΔ We have 21 ∠=∠ (Vertically opposite angles) ∠ALO ∠L = ∠DMO = 90° Therefore, DMO~ALO ΔΔ (By AA similarity criterion)

⇒ DOAO

DMAL

= ...(i)

Now, DMBC

21

ALBC21

DBC)( of AreaABC)( of Area

××

××=

ΔΔ

From equation (i), We have

DMAL

DBC)( of AreaABC)( of Area

=ΔΔ

Hence, Area of ( ABC) AL AOArea of ( DBC) DM DO

Δ= =

Δ

Q.4. If the areas of two similar triangles are equal, prove that they are congruent.

Solution. Given: PQR and ABC ΔΔ are similar and equal in area. To Prove: PQRABC Δ≅Δ Proof: Since, PQR~ABC ΔΔ

( )( )

2

2Area of ABC BCTherefore, Area of PQR QR

Δ=

Δ

⇒ 1QRBC

2

2

= [Since, Area (ΔABC) = (ΔPQR)]

⇒ BC2 = QR2 ⇒ BC = QR

Similarly, we can prove that, AB = PQ and AC = PR Therefore, ΔABC ≅ ΔPQR (By SSS Criterion of congruence)

[If two triangles are similarcorresponding sides are proportional]

1[Area of triangle = × Base × Altitude]2

24

Q.5. D, E and F are respectively the mid-points of the sides AB, BC and CA of ΔABC. Determine the ratio of the areas of ΔDEF and ΔABC.

Solution. Given: D, E and F are the mid-points of the sides AB, BC and CA respectively of the ABC.Δ

To find: area DEF)(Δ and area ABC)(Δ Proof: In ABC,Δ We have F is the mid-point of AB (Given) E is the mid-point of AC (Given) So, by the mid-point theorem, We have

BC21FE and BC||FE =

⇒ BD||FE and BDFE = BC]21BD [ =∵

∴ BDEF is a || gm. [ Opp. sides of a || gm are parallel and || equal]∵ Similarly in FBD and DEF,Δ Δ We have DEFB = (Opposite sides of || gm BDEF) FDFD = (Common) FEBD = (Opposite sides of || gm BDEF) ∴ DEFFBD Δ≅Δ (SSS Congruence Theorem) Similarly, we can prove that: DEFAFE Δ≅Δ DEFEDC Δ≅Δ If triangles are congruent, then they are equal in area. So, DEF)( areaFBD)( area Δ=Δ ...(i) DEF)( areaAFE)( area Δ=Δ ...(ii) and, DEF)( areaEDC)( area Δ=Δ ...(iii)

Now, area ( ABC) area ( FBD) area ( DEF)Δ = Δ + Δ area ( AFE) area ( EDC)+ Δ + Δ ...(iv)

area ( ABC) area ( DEF) area ( DEF) area ( DEF) area ( DEF)Δ = Δ + Δ + Δ + Δ

⇒ ABC)( area41DEF)( area Δ=Δ [From (i), (ii) and (iii)]

⇒ 41

ABC)( areaDEF)( area

=ΔΔ

Hence, 4:1ABC)( area : DEF)( area =ΔΔ

25

Q.6. Prove that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding medians.

Solution. Given: AM and DN are the medians of triangles ABC and DEF respectively

and, ABC ~ DEFΔ Δ

To Prove: 2

2

DNAM

DEF)( areaABC)( area

=ΔΔ

Proof: DEF~ABC ΔΔ (given)

Therefore, 2

DEAB

DEF)( areaABC)( area

⎟⎠⎞

⎜⎝⎛=

ΔΔ ...(i)

and, FDCA

EFBC

DEAB

== ...(ii)

⇒

1 BCAB CD2 1DE FDEF2

= =

In DEN and ABM ΔΔ

EB ∠=∠ [Since, ΔABC ~ ΔDEF]

ENBM

DEAB

= [Prove in (i)]

Therefore, ABC ~ DEFΔ Δ [By SAS criterion of similarity]

Therefore, DNAM

DEAB

= ...(iv)

Therefore, ΔABM ~ ΔDEN

As the areas of two similar triangles are proportional to the squares of the corresponding sides.

Therefore, 2

2

2

2

DNAM

DEAB

DEF)( areaABC)( area

==ΔΔ

Hence proved.

Q.7. Prove that the area of an equilateral triangle described on one side of a square is equal to half the area of the equilateral triangle described on one of its diagonals.

26

Solution. Given: ABCD is a square whose one diagonal is AC. BQC and APC ΔΔ are two equilateral triangles described on the diagonals AC and side BC of the square ABCD.

To Prove: APC)( area21BQC)( area Δ=Δ

Proof: BQC and APC ΔΔ are both equilateral triangles. (Given)

Therefore, BQC~APC ΔΔ [AAA similarity criterion]

Therefore, 2

22

BCAC

BCAC

BQC)( areaAPC)( area

=⎟⎠⎞

⎜⎝⎛=

ΔΔ

⇒ 2BCBC2

BCBC2

2

22

==⎟⎟⎠

⎞⎜⎜⎝

⎛ [Since, Diagonal 2 side 2 BC]= =

= BQC)( area2APC)( area Δ×=Δ

or, APC)( area21BQC)( area Δ=Δ

Hence proved.

Q.8. Tick the correct answer and justify: ABC and BDE are two equilateral triangles such that D is the mid-point of BC.

Ratio of the areas of triangles ABC and BDE is (A) 2 : 1 (B) 1 : 2 (C) 4 : 1 (D) 1 : 4.

Solution. BDE and ABC ΔΔ are two equilateral triangles. D is the mid-point of BC.

Therefore, BC21DCBD ==

Let each side of triangle is 2a. As, BDE~ABC ΔΔ

Therefore, 1:4144

)()2(

BDAB

BDE)( areaABC)( area

2

2

2

2

2

2

=====ΔΔ

aa

aa

Thus, correct option is (C).

Q.9. Sides of two similar triangles are in the ratio 4 : 9. Areas of these triangles are in the ratio (A) 2 : 3 (B) 4 : 9 (C) 81 : 16 (D) 16 : 81

27

Solution. Let ABC and DEF are two similarly triangles DEF~ABC ΔΔ (Given)

and, 94

EFBC

DFAC

DEAB

=== (Given)

Therefore, 2

2

DEAB

DEF)( areaABC)( area

=ΔΔ

[The ratio of the areas of two triangles is equal to square of the ratio of their corresponding sides]

Therefore, 81:168116

94

DEF)( areaABC)( area 2

==⎟⎠⎞

⎜⎝⎛=

ΔΔ

Thus, correct option is (D).

Exercise 6.5 Q.1. Sides of triangles are given below. Determine which of them are right triangles.

In case of a right triangle, write the length of its hypotenuse. (i) 7 cm, 24 cm, 25 cm (ii) 3 cm, 8 cm, 6 cm (iii) 50 cm, 80 cm, 100 cm (iv) 13 cm, 12 cm, 5 cm.

Solution. In a right triangle the square of the hypotenuse is equal to the sum of squares of the other two sides.

(i) 7 cm, 24 cm, 25 cm Let 7=a cm, 24=b cm, 25=c cm As, 625)25(;576)24(;49)7( 222 === We have, a2 + b2 = (7)2 + (24)2 = 252 = c2 Hence, the given triangle is right angled. Length of hypotenuse 25= cm.

(ii) 3 cm, 8 cm, 6 cm Let 3=a cm, 6=b cm, 8=c cm As, 36)6(;64)8(;9)3( 222 ===

We have, a2 + b2 = 32 + 62 = 45 ≠ c2

2 2 28 3 6≠ + ⇒ 456436964 ≠⇒+= Hence, the given triangle is not right angled.

(iii) 50 cm, 80 cm, 100 cm Let 100=c cm, 80=b cm, 50=a cm Hence, the given triangle is not right angled.

28

(iv) 13 cm, 12 cm, 5 cm Let 13=c cm, 12=a cm, 5=b cm

2 2 213 169;12 144; 5 25= = =

Here, 222 bac += We have, a2 + b2 = (80)2 + (50)2 = 6400 + 2500 = 8900 ≠ c2 ⇒ 222 )12()5()13( += ⇒ 16914425169 =+= Hence, the given triangle is right angled. Length of the hypotenuse = 13 cm

Q.2. PQR is a triangle right angled at P and M is a point on QR such that QR.PM ⊥ Show that MR.QMPM 2 ⋅=

Solution. Given: PQRΔ is right angled at P and M is a point on QR such that QR.PM ⊥

To Prove: MR.QMPM2 ⋅=

Proof: In ΔPQM, We have PQ2 = PM2 + QM2 [By Pythagoras theorem] Or, PM2 = PQ2 – QM2 ….(i) In ΔPMR, We have PR2 = PM2 + MR2 [By Pythagoras theorem] Or, PM2 = PR2 – MR2 …(ii)

Adding (ii) and (ii), We get 2PM2 = (PQ2 + PM2) – (QM2 + MR2)

= QR2 – QM2 – MR2 [∴ QR2 = PQ2 + PR2] = (QM + MR)2 – QM2 – MR2 = 2QM, MR

Therefore, PM2 = QM, MR, Hence proved.

Q.3. In the given figure, ABD is a triangle right angled at A and AC BD.⊥ Show that

(i) ⋅2AB = BC BD (ii) ⋅2AC = BC DC

(iii) ⋅2AD = BD CD Solution. Given: A right angled ABDΔ right angled at A and BD.AC ⊥

To Prove: (i) BDBCAB2 ⋅=

29

(ii) DCBCAC2 ⋅= (iii) CDBDAD2 ⋅= Proof: In DCA, and DAB ΔΔ We have DD ∠=∠ (Common) ∠BAD = ∠DCA = 90° Therefore, DCA~DAB ΔΔ ...(i) [By AA similarity] In ACB, and DAB ΔΔ BB ∠=∠ (Common) ∠BAD = ∠BCA = 90° Therefore, ACB~DAB ΔΔ ...(ii) From equation (i) and equation (ii), We have DCA,~ACB~DAB ΔΔΔ

(i) DAB~ACB ΔΔ (proved above)

Therefore, 2

2

DBAB

DAB)( areaACB)( area

=ΔΔ

⇒ 2

2

DBAB

ACDB21

ACBC21

=×

×

⇒ 2

2

DBAB

DBBC

=

⇒ DBAB

1BC 2

=

⇒ BDABBC

2

=

Therefore, BDBCAB2 ×= (ii) DCA~ACB ΔΔ (proved above)

Therefore, 2

2

DCAC

DCA)( areaACB)( area

=ΔΔ

⇒ 2

2

1 BC AC AC21 DCDC AC2

×=

×

⇒ 2

2BC ACDC DC

=

30

⇒ DCACBC

2

=

Therefore, DCBCAC2 ×= (iii) DCA~DAB ΔΔ (proved above)

2

2

DBDA

DCA)( areaDAB)( area

=ΔΔ

⇒ 2

2

1 CD AC AD21 BDBD AC2

×=

×

⇒ 2

2CD ADBD BD

=

⇒ BD

ADCD2

=

Therefore, CDBDAD2 ×=

Q.4. ABC is an isosceles triangle right angled at C. Prove that .2ACAB 22 =

Solution. Given: ABCΔ is an isosceles triangle right angled at C. To Prove: .AC2AB 22 = Proof: In °=∠Δ 90C ACB, BCAC = (Given) 222 BCACAB += [By using Pythagoras theorem] 22 ACAC += [Since, BC = AC] 22 AC2AB = Hence, proved.

Q.5. ABC is an isosceles triangle with AC = BC. If 2 2AB = 2AC , prove that ABC is right triangle.

Solution. Given: ABC is an isosceles triangle having AC = BC and 22 AC2AB =

To prove: ABCΔ is a right triangle. Proof: 22 AC2AB = (given) ⇒ 222 ACACAB += Or, 222 BCACAB += [Since, AC = BC] Hence, (By Pythagoras Theorem) ABCΔ is right angled triangle.

31

Q.6. ABC is an equilateral triangle of side 2a. Find each of its altitudes.

Solution. ABC is an equilateral triangle of side a2 . Draw, BCAD ⊥

In ΔADB and ΔADC, We have AB = AC [Given] AD = AD [Common] ∠ADB = ∠ADC [Each 90°]

Therefore, ADCADB Δ≅Δ [By RHS congruence] Hence, a== DCBD [CPCT]

In right angled ADBΔ

3AD

3AD AD4

)(AD)2( BDADAB Here,

22

222

222

222

a

aaa

aa

=⇒

=⇒

=−⇒

+=⇒

+=

Q.7. Prove that the sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals.

Solution. Given: ABCD is a rhombus whose diagonals AC and BD intersect each other at O.

To Prove: 222222 BDACADCDBCAB +=+++ Proof: Since, the diagonals of a rhombus bisect each

other at right angles. Therefore, DOBO and COAO == In °=∠Δ 90AOB AOB,

Therefore, 222 BOAOAB += ...(i) [By Pythagoras Theorem] Similarly, 222 BOCOBC += ...(ii) 222 ODCOCD += ...(iii) 222 AODODA += ...(iv) Adding, equation (i), equation (ii), equation (iii) and equation (iv), we have

2222

22

22222222

BDACBO)2((2AO)DO]BO and COAO Since,[BO4AO4

DO2BO2CO2AO2DACDBCAB

+=+=

==+=

+++=+++

32

Q.8. In the given figure, O is a point in the interior of a triangle ABC, OD BC, OE AC⊥ ⊥ and OF AB.⊥ Show that

(i) − − −2 2 2 2 2 2 2 2 2OA + OB + OC OD OE OF = AF + BD + CE

(ii) 2 2 2 2 2 2AF + BD + CE = AE + CD + BF

Solution. Given: ABCΔ in which AB.OF and ACOE BC,OD ⊥⊥⊥

To Prove: (i) 222222222 OFOEODOCOBOACEBDAF −−−++=++ (ii) 222222 BFCDAECEBDAF ++=++ Proof: (i) In right angled AFO,Δ we have

222 AFOFOA += [By Pythagoras Theorem] ⇒ 22 OFOAAF −= 2 ...(i) In right angle BDO,Δ we have,

222 ODBDOB += [By Pythagoras Theorem] ⇒ 222 ODOBBD −= ...(ii) In right angle CEO,Δ we have,

222 OECEOC += [By Pythagoras Theorem] ⇒ 222 OEOCCE −= ...(iii) Adding equation (i), equation (ii) and equation (iii), we have AF2 + BD2 + CE2 222222 OFOEODOCOBOA −−−++= Which proves part (i),

222

222222222

BFCDAE)OFOB()ODOC()OE(OACEBDAF Again,

++=

−+−+−=++

]OFOBBF ;ODOCCD ;OEAOAE[ 222222222 −=−=−=

33

Q.9. A ladder 10 m long reaches a window 8 m above the ground. Find the distance of the foot of the ladder from base of the wall.

Solution. Height of window from ground m 8(AB) = . Length of ladder m 10AC)( = Distance between foot of ladder and foot of wall ?BC)( = In ABC,Δ We have

222 ACBCAB =+ [By Pythagoras Theorem]

2 2 2

2

2

(8) (BC) (10)

64 BC 100

BC 100 64

BC 36 BC 6 m

⇒ + =

⇒ + =

⇒ = −

⇒ =⇒ =

Therefore, Distance between foot of ladder and base of the wall 6= m.

Q.10. A guy wire attached to a vertical pole of height 18 m is 24 m long and has a stake attached to the other end. How far from the base of the pole should the stake be driven so that the wire will be taut?

Solution. Height of pole (AB) 18= m Length of wire (BC) 24= m In right angle triangle ABC, We have AB2 + BC2 = AC2 [By Pythagoras Theorem] ⇒ (18)2 + (BC)2 = (24)2 ⇒ 324 + (BC)2 = 576 ⇒ BC2 = 576 – 324

⇒ BC 252= ⇒ BC = 76

Hence, the stake should be driven by 76 m from the base of the pole so that the wire will be taut.

Q.11. An aeroplane leaves an airport and flies due north at a speed of 1000 km per hour. At the same time, another aeroplane leaves the same airport and flies due west at a speed of 1200 km per hour. How far apart will be the two planes after

112

hours?

Solution. Speed of first aeroplane 1000= km/hr.

34

Distance covered by first aeroplane due north in 211 hours (OA)

231000 ×= km

= 1500 km Speed of second aeroplane 1200= km/hr.

Distance covered by second aeroplane due west in 211 hours (OB)

231200 ×= km

Or, km 1800OB = In right angle AOB,Δ We have AB2 = AO2 + OB2 ⇒ AB2 = (1500)2 + (1800)2

⇒ AB = 2250000 3240000+

= 5490000

Or, AB = 300 61 km

Hence, distance between two aeroplanes will be 300 61 km

Q.12. Two poles of heights 6 m and 11 m stand on a plane ground. If the distance between the feet of the poles is 12 m, find the distance between their tops.

Solution. Height of first pole (AB) = 11 m Height of second pole (CD) = 6 m Distance between foot of pole (BD) = 12 m As, m 6DCBE == Therefore, BEABAE −= = (11 – 6) m = 5 m In right angle AEC,Δ We have

2 2 2

2 2

AC AE EC

AC (5) (12) 25 144 169 13.

= +

= + = + = =

Hence, distance between their top m. 13=

Q.13. D and E are points on the sides CA and CB respectively of a triangle ABC right angled at C. Prove that 2 2 2 2AE + BD = AB + DE .

Solution. Given: ABCΔ is right angled at C, D and E are the mid-points of CA and BC respectively.

35

i.e., AC21ADCD ==

and, BC21ECBE ==

To Prove: 2222 DEABBDAE +=+ Proof: In right angles BCA,Δ We have

222 CABCAB += ...(i) [By Pythagoras Theorem] In right angle ECD,Δ We have

222 DCECDE += ...(ii) [By Pythagoras Theorem] In right angled ACE,Δ We have

222 CEACAE += ...(iii) [By Pythagoras Theorem] In right angled BCD,Δ We have

222 CDBCBD += ...(iv) [By Pythagoras Theorem] Adding equation (iii) and equation (iv), we have

22

2222

222222

DEAB]DCCE[]CBAC[

CDBCCEACBDAE

+=

+++=

+++=+

Hence, .DEABBDAE 2222 +=+ Which is the required result.

Q.14. The perpendicular from A on side BC of a ΔABC intersects BC at D such that DB = 3CD. (See figure). Pove that .BC2AC2AB 222 +=

Solution. Given: ABC, AD BC,Δ ⊥ 3CD.BD =

To Prove: 222 BC2ACAB2 += Proof: In right angles triangles ADB and ADC, we have 222 BDADAB += ...(i) 222 DCADAC += ...(ii) [By Pythagoras Theorem]

Subtracting equation (ii) from equation (i), we get 2222 DCBDACAB −=−

22

22

4BC88CD

3CD]BD [CDCD9

⎟⎠⎞

⎜⎝⎛==

=−= ∵

4CD]CD3CDCDDBBC Since,[ =+=+=

36

Therefore, 2

BCACAB2

22 =−

⇒ 222 BC)ACAB(2 =−

⇒ 222 BCAC2AB2 =− ∴ 222 BCAC22AB += Which is the required result.

Q.15. In an equilateral triangle ABC, D is a point on side BC such that 1BD = BC.3

Prove that 2 29AD = 7AB . Solution. Given: Equilateral triangle ABC, D is a point on side BC such that

1BD BC.3

=

To Prove: 22 AB7AD9 = Proof: In ΔABC, We have

BM = MC = 12

BC

OM = BM – BD

= 12

BC – 13

BC [As, BD = 13

BC]

= BC6

… (ii)

Now, In ΔADM, We have AD2 = AM2 + DM2 [By Pythagoras Theorem] But, In ΔAMC, We have AM2 = AC2 – MC2 [By Pythagoras Theorem] Therefore, AD2 = AC2 + DM2 – MC2

= AC2 + 2 2BC BC –

6 2⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

[From (i) and (ii)]

= AC2 +

2 2BC BC36 4

−

= AB2 +

2 2AB AB36 4

− [Since, AB = BC = AC]

= 21 11 AB

36 4⎛ ⎞+ −⎜ ⎟⎝ ⎠

[Since, ABC is an equilateral triangle and AM BC] ... (i)⊥

37

= 27 AB

a ⇒ 9AD2 = 7AB2, Hence, proved.

Q.16. In an equilateral triangle, prove that three times the square of one side is equal to four times the square of one of its altitudes.

Solution. Given: ABC is an equilateral triangle and AD ⊥ BC To Prove: 22 AD4AB3 = Proof: In ABC,Δ Let a2ACBCAB === Also, BCAD ⊥

Therefore, a=== BC21DCBD

In ABDΔ , We have AB2 = AD2 + BD2 ⇒ (2a)2 = AD2 + (a)2 ⇒ 4a2 = AD2 + a2 ⇒ 4a2 – a2 = AD2

2

2 2 ABAD = 3 = 3 2

a ⎛ ⎞⎜ ⎟⎝ ⎠

[Since, AB = 2a]

⇒ AD2 = 3 2AB

4

⇒ 3AB2 = AD2 Which is the required result.

Q.17. Tick the correct answer and justify: In ΔABC, AB = 6 3 cm, AC = 12 cm and BC = 6 cm. The angle of B is:

(A) 120° (B) 60° (C) 90° (D) 45°

Solution. Here, cm 12AC =

cm 36AB = and, cm 6BC =

Now, 36108)6()36(BCAB 2222 +=+=+ = 144 = AC2 Hence by converse of pythagoras theorem ABCΔ is right angled triangle right

angled at B. Therefore, °=∠ 90B

38

Exercise 6.6 (Optional) Q.1. In the given figure, PS is the bisector of ∠QPR of ΔPQR. Prove that

QS PQ= .SR PR

Solution. Given: PS PQR,Δ is the bisector of QPR.∠ i.e., 21 ∠=∠

To Prove: PRPQ

SRQS

=

Condition : Draw RT Parallel to PS. Entend PQ to meet RT at T.

Proof: In QRT,Δ We have TR||PS Therefore, 32 ∠=∠ (Alternate angles) Therefore, 41 ∠=∠ (Corrresponding angles) Therefore, 21 ∠=∠ (Given) Therefore, 43 ∠=∠ In PRT,Δ We have 43 ∠=∠ Therefore, PRPT = As, TR||PS

Therefore, SRQS

PTQP

= [By Basic Proportionality Theorem]

⇒ SRQS

PRQP

= [Since, PT PR]=

⇒ SRQS

PRPQ

=

Which is the required result.

Q.2. In the given figure, D is a point on hypotenuse AC of ΔABC, DM BC⊥ , DN ⊥ AB, prove that:

(i) ⋅2DM = DN MC (ii) ⋅2DN = DM AN

Solution. Given: ABDN and BCDM ABC, ⊥⊥Δ

To Prove: (i) ACDNDM2 ⋅= (ii) AMDMDN2 ⋅= Proof: In ΔABC, ACBD ⊥ (Given) ⇒ °=∠ 90BDC

Side opposite to equal anglesof a triangle are equal

⎡ ⎤⎢ ⎥⎣ ⎦

39

⇒ °=∠+∠ 90MDCBDM ...(i) In DMC,Δ We have °=∠ 90DMC ⇒ °=∠+∠ 90MDCC ...(ii) From equation (i) and equation (ii), we have MDCCMDCBDM ∠+∠=∠+∠ ⇒ CBDM ∠=∠ Now in MDC, and BMD ΔΔ We have CBDM ∠=∠ [Proved above] °=∠=∠ 90DMCBMD Therefore, MDC~BMD ΔΔ [By AA criterion of similarity]

⇒ DMMC

BMDM

=

⇒ MCBMDM2 ×= ⇒ MCDNDM2 ⋅= [Since, BM DN]= Similarly, in NBD~NDA ΔΔ , We have

⇒ DNAN

BNDN

=

⇒ ANBNDN2 ×= ⇒ ANDMDN2 ⋅= Hence proved

Q.3. In the given figure, ABC is a triangle in which ∠ ⊥ABC > 90° and AD BC produced. Prove that ⋅2 2 2AC = AB + BC + 2BC BD .

Solution. Given: BCAD ABC, ⊥Δ when produced, .90ABC °>∠

To Prove: BDBC2BCABAC 222 ⋅++= Proof: In ADCΔ , We have 222 DC)(AD)((AC) += ...(i)

⇒ 2 2 2(AC) (AD) (BC BD)= + +

⇒ BD2BCBDBCADAC 2222 ⋅+++=

2 2AB BC 2BC BD= + + ⋅ (Since, )BDADAB 222 += Hence Proved.

Q.4. In the given figure, ABC is a triangle in which ∠ ⊥ABC < 90° and AD BC. prove that

− ⋅2 2 2AC = AB + BC 2BC BD

[Since, corresponding sides of similar triangles are proportional]

[Since, corresponding sides of similar triangles are proportional]

40

Solution. Given: ABC, ABC 90 and AD BC.Δ ∠ < ° ⊥

To Prove: BDBC2BCABAC 222 ⋅−+= Proof: ADCΔ is right-angled at D Therefore, 222 DACDAC += ...(i) [By Pythagoras Theorem] Also, ΔADB is right angled at D Therefore, 222 DBADAB += [By Pythagoras Theorem] ...(ii) From equation (ii), We have

2 2 2AD AB DB= −

⇒ 2 2 2 2AC CD (AB BD )= + − [From equation (i)

⇒ 2 2 2 2AC (BC BD) AB BD= − + − (Since, BC BD DC)= +

⇒ 2 2 2 2 2(AC) BC BD 2BC BD AB BD= + − ⋅ + −

⇒ BDBC2BCABAC 222 ⋅−+= , Hence Proved.

Q.5. In the given Figure, AD is a median of a triangle ABC and ⊥AM BC. Prove that:

(i) ⎛ ⎞⎜ ⎟⎝ ⎠

22 2 BCAC = AD + BC. DM +

2

(ii) ⎛ ⎞− ⎜ ⎟⎝ ⎠

22 2 BCAB = AD BC. DM +

2

(iii) 2 2 2 21AC + AB = 2AD + BC2

Solution. (i) In AMC,Δ We have

222 MC)(AM)((AC) += [By Pythagoras Theorem]

⇒ 222 DC)MD((AM)(AC) ++= [Where, DC]MDMC +=

⇒ DCMD2(DC)(MD)AM)((AC) 2222 ⋅+++=

In 2 2 2AMD, AD AM MDΔ = +

⇒ DCMD22

BCAD)((AC)2

22 ⋅+⎟⎠⎞

⎜⎝⎛+=

⇒ DMBC2

BCAD(AC)2

22 ⋅+⎟⎠⎞

⎜⎝⎛+= Proved [Since, DC = BD]

(ii) In ΔAMB, We have

2 2 2In AMD, AD = AM + MD1and DC = BC2

⎡ ⎤Δ⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

41

222 MBAMAB += [By Pythagoras Theorem] ⇒ 222 MD)BD(AMAB −+=

⇒ 2 2 2 2AB AM (BD) (MD) 2BD. MD= + + −

2 2 2AM MD BD 2BD MD= + + − ⋅

⇒ 2

2 2 BC BCAB AD 2 MD2 2

⎛ ⎞= + − × ⋅⎜ ⎟⎝ ⎠

⇒ MDBC2

BCADAB2

22 ⋅−⎟⎠⎞

⎜⎝⎛+=

⇒ 2

22

2BCMDBCADAB ⎟

⎠⎞

⎜⎝⎛+⋅−= , Hence Proved

(iii) Adding the result obtained in (i) and (ii), We get

2

2 2 2 BCAC AB 2(AD) 22

⎛ ⎞+ = + ⎜ ⎟⎝ ⎠

⇒ 4

BC2AD2ABAC2

222 ×+=+

⇒ 2

BCAD2ABAC2

222 +=+

Hence Proved.

Q.6. Prove that the sum of the squares of the diagonals of a parallelogram is equal to the sum of squares of the sides.

Solution. Given: ABCD is a parallelogram whose diagonals are AC and BD. To Prove: 222222 (BD)(AC)(DA)(CD)(BC)(AB) +=+++ Proof : In AMD,Δ We have BCAD = [Opposite sides of a parallelogram] BNAM = [Both are altitudes of the same parallelogram to the same base] Therefore, BNCAMD Δ≅Δ [By RHS congruence criterion] Construction : Draw AM ⊥ CD and BN ⊥ CD. Hence, NCMD = ...(i) [CPCT] In BND,Δ We have

222 DN)(BN)((BD) += [By Pythagoras Theorem]

2 2 2 BC[Since, AD AM MD and BD ]2

= + =

42

22 CN)DC(BN ++=

2 2 2BN DC CN 2DC CN= + + + ⋅

2 2 2BN CN DC 2DC CN= + + + ⋅

Or, 2 2 2(BD) (BC) (DC) 2DC CN= + + ⋅ ...(ii) [Since, 222 CN)((BN)(BC) += ] In AMC,Δ We have

2 2 2

2 2

2 2 2

2 2 2

2 2 2

(AC) AM MC

AM (DC DM)

AM DC DM 2DC DM

AM DM DC 2DC DM

Or, (AC) AD DC 2DC DM

= +

= + −

= + + −

= + + −

= + −

⋅

⋅

⋅

[Where, ]DMAMAD 222 += Or, CN2DCABADAC 222 ⋅−+= ...(iii) [Since, AB = DC] Adding equation (ii) and equation (iii), we have

2222

222222

DACDBCAB)DCBC()AB(ADBDAC

+++=

+++=+

⇒ 222222 (DA)(CD)(BC)(AB)BD)((AC) +++=+ Hence Proved.

Q.7. In the given figure, two chords AB and CD intersect each other at the point P. Prove that : (i) ΔAPC ~ ΔDPB (ii) ⋅⋅AP PB = CP DP

Solution. Given: AB and CD are two chords of a circle intersects each other at P. To Prove: (i) DPB~APC ΔΔ (ii) AP PB CP DP=⋅ ⋅ Proof: (i) In DPB, and APC ΔΔ We have DPBAPC ∠=∠ (Vertically opposite angle) CABCDB ∠=∠ (Angle in the same segment) Therefore, DPB~APC ΔΔ (By AA similarity criterion)

(ii) As, DPB~APC ΔΔ [Proved in part (i)]

Therefore, PBPC

DPAP

=

⇒ DPPCPBAP ⋅=⋅ Hence Proved.

( )If two triangles are similar corresponding sides are proportional

43

Q.8. In the given figure, two chords AB and CD of a circle intersect each other at point P (when produced) outside the circle. Prove that: (i) ΔPAC ~ ΔPDB (ii) ⋅ ⋅PA PB = PC PD

Solution. Given: AB and CD are two chord of circle intersects each other at P (when produced) out side the circle.

To Prove: (i) PDB~PAC ΔΔ

(ii) PDPCPBPA ⋅=⋅

Proof: (i) In PDB and PAC ΔΔ , We have

PBDPCA ∠=∠

Exterior angle of a cyclic quadrilateral is equal to interior opposite angle)

Also, ∠P = ∠P [Common]

Therefore, PDB~PAC ΔΔ (AA similarity criterion)

(ii) As, PDB~PAB ΔΔ

Therefore, PBPC

PDPA

=

⇒ PDPCPBPA ×=×

PDPCPBPA ⋅=⋅

Q.9. In the given figure, D is a point on side BC of ΔABC

such that BD AB= .CD AC

Prove that AD is the bisector of ∠BAC.

Solution. Given: D ABC,Δ is a point on BC such that

ACAB

DCBD

=

To Prove: AD is bisector of BAC,∠ i.e., CADBAD ∠=∠ Construction : Draw CE || AD to meet AB, (when extended at E) Proof: In BCE,Δ we have

[ ]We know if two triangle are similar corresponding sides are proportional

44

CE||AD ...(By construction) So, By Basic Proportionality Theorem, We have

AB AB

BD ABAE AC

DC AE AE AC

⇒ =⎫= ⎬⎭ ⇒ =

BD ABSince, CD AC

⎡ ⎤=⎢ ⎥⎣ ⎦

In ACE,Δ we have: ACAE = [Proved, above] ⇒ ACEAEC ∠=∠ ...(i) Also, ACECAD ∠=∠ [Since, AD || CE] ...(ii) Similarly, AECBAD ∠=∠ [Since, AD || CE] ...(iii) Using equation (i), equation (ii) and equation (iii), we have CADBAD ∠=∠ Hence Proved.

Q.10. Nazima is fly fishing in a stream. The tip of her fishing rod is 1.8 m above the surface of the water and the fly at the end of the string rests on the water 3.6 m away and 2.4 m from a point directly under the tip of the rod. Assuming that her string (from the tip of her rod to the fly) is taut, how much string does she have out (see figure)? If she pulls in the string at the rate of 5 cm per second, what will the horizontal distance of the fly from her after 12 seconds?

Solution. In triangle ABC, We have m 1.8AB = m 2.4BC = °=∠ 90B

2 2 2

2 2 2

2

2 2

Therefore, AC AB BC (By Pythagoras Theorem)

AC (1.8) (2.4)

AC 3.24 5.76 9

AC (3) AC 3 m

= +

⇒ = +

⇒ = + =

⇒ =⇒ =

Length of the string pulled in 12 second 5 12 60 cm 0.6 m= × = =

Length of remaining string left out. m )6.0(3AD −=

45

m 2.4AD = In ABD,Δ We have

222 BDABAD += [By Pythagoras Theorem] ⇒ 222 ABADBD −= ⇒ 2 2 2(BD) (2.4) (1.8)= −

⇒ 2BD 5.76 3.24= −

⇒ 2BD 2.52 m= ⇒ BD 1.587 m=

Therefore, Horizontal distance of the fly from Nazima

m 2.79m 2.787

m 1.2)(1.587m 1.2BD

==

+=+=

Hence, original length of string and horizontal distance of the fly from Nazima is 3 m and 2.79 m respectively.