the equidistance theorems advanced geometry chapter 4.4

The EQUIDISTANCE The EQUIDISTANCE TheoremsTheorems

The EQUIDISTANCE The EQUIDISTANCE TheoremsTheorems

Advanced Geometry Chapter 4.4Advanced Geometry Chapter 4.4

Some basics:

A.) Distance – the length of the path between two objects.

B.) A line segment is the shortest path between two points.

segment itselfAB

length of segmentAB

Definition

Postulate

A

B

What does Equidistance look like?

If 2 points P and Q are the same distance from a third point X, then X is said to be _____________from P and Q.

EQUIDISTANT

Q

P

XX

Definition of Perpendicular Bisector:

A perpendicular bisector of a segment is the line that both _________ and is _____________ to the segment.

BISECTSPERPENDICUL

AR

A B

L

X

is bisector ofL AB

Theorem 24: If 2 points are equidistant from the endpoints of a segment,

NEEDED: 2 points equidistantfrom endpoints of same

segment, or

then the two points DETERMINE the PERPENDICULAR BISECTOR of the segment. (Perpendicular Bisector Theorem)

2 pairs of congruent segments

A C

6) PBT (Perpendicular Bisector Theorem)

AA AA

CCCC

Given: ΔABC ≅ ΔADC

Prove: AC bis of BDA E

C

D

B

1) ΔABC ≅ ΔADC 1) Given

2) AB ≅ AD 2) CPCTC

AA3) A is =dist from B & D 3) If 2≅segs, then a point is =dist

from endpts of another seg4) CB ≅ CD

CC

4) CPCTC

5) C is =dist from B & D 5) Same as #3

6) AC is bis of BD 6) If 2 points are =dist from endpts of same seg, then

they determine the perpendicular bisector of the seg. (3, 5)

Prove: AL AO

Theorem 25: If a point is ON the perpendicular bisector of a segment, then . . .

NEEDED: Perpendicular Bisector of the Segment.

. . . the point is EQUIDISTANT from the endpoints of the segment. (Perpendicular Bisector Theorem)

OL

Given: HI is bisector of LO

C

B

AI

H

Prove: IL IO Prove: BL BO

Prove: HL HO

Given:Prove:

1.

2.

3. X is =dist from A & B

4. 5. F is =dist from A & B6. 7.

1. Given

2. Reflexive

3.If a point is on the perp bisector, then it is =dist. (1)

4. A bis seg into 2 ≅ segs (3)5. Same as #3 (1)6. Same as #4 (5)7. SSS (2,4,6)

Bisector ofEF ABBFXAFX

A

FX

B

E

A

FX

B

E

ABofBisEF

XFXF

XBXA

FBFA BFXAFX

S

S

S

3. PBT (1)

FX

TRUE/FALSEPRACTICE

Ready??

C

E

B

DA

BCoftorbiAD sec

TRUE

TRUE

OR

FALSE

???

Given:

1. E is the midpoint of BC.

C

E

B

DA

BCoftorbiAD sec

2. ∡AEC is a right angle

TRUE

Given:

TRUE

OR

FALSE

???

C

E

B

DA

BCoftorbiAD sec

FALSE

3. E is the midpoint of AD

Given:

TRUE

OR

FALSE

???

C

E

B

DA

BCoftorbiAD sec

TRUE

4. AC ≅ AB

Given:

TRUE

OR

FALSE

???

C

E

B

DA

BCoftorbiAD sec

TRUE

5. CE ≅ BE

Given:

TRUE

OR

FALSE

???

C

E

B

DA

BCoftorbiAD sec

FALSE

6. CA ≅ CD

Given:

TRUE

OR

FALSE

???

C

E

B

DA

BCoftorbiAD sec

FALSE

7. AE ≅ ED

Given:

TRUE

OR

FALSE

???

C

E

B

DA

BCoftorbiAD sec

FALSE

8. CB bisects AD

Given:

TRUE

OR

FALSE

???

ProveProve the following statement:

The line drawn from the vertex angle of an isosceles triangle

through the point of intersection of the medians to the legs is perpendicular to the base.

Given: Given: ΔΔABC is isosceles, base BCABC is isosceles, base BC BF is median to side ACBF is median to side AC

CE is median to side AB CE is median to side AB Prove: AD is the altitude to BCProve: AD is the altitude to BC

DD

AA

CCBB

EE FFSince ABC is isosceles with base BC, AB and AC are congruent legs. This also gives us congruent base angles, ∡ABC and ∡ACB.BF and CE are medians to the legs, AC and AB, respectively. This means that EB and FC are congruent by division property. BC is a reflexive side, so Δ EBC and ΔFCB are congruent by SAS. We now have proven ∡PBC and ∡PCB congruent by CPCTC.Now, ΔBPC is isosceles, with base BC and congruent legs PB and PC.

Now with PB = PC, and we know that PD is a reflexive side of congruent Δ’s DPB and DPC (∡PDB and ∡PDC are supplementary and adjacent, which means right angles.Right ∡’s ΔDPB ΔDPC by HL)Then, Right ∡s segs, ∴ AD is the altitude to BC !

PPOr, we could say now that because of our isosceles

triangles giving us 2 pairs of are congruent segs, two points (A and P) are equidistant from the endpoints of segment (BC),

so they determine the PERPENDICULAR bisector of the

segment by PBT. Now, since D is ON line AP, by

relying upon perpendicular bisector theorem again we can

conclude that AD is the altitude to BC!

D

Which method do you

prefer???

4.4 Homework4.4 Homework

Pp 187 – 190

(2 – 4; 7 – 10; 12, 15, 18, 19)