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Preparing for the Regents Examination Geometry, AK

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3. −−<br />

AB −−<br />

AB<br />

4. −−<br />

AC −−<br />

BD<br />

5. ABC BAD<br />

2 1. 1 3<br />

2. 2 2<br />

3. 1 3 2 3<br />

(SAS SAS)<br />

4. ADB EDC<br />

5. 4 5<br />

6.<br />

(Partition<br />

postulate)<br />

−−−<br />

AD −−<br />

ED<br />

7. ADB EDC (ASA ASA)<br />

3 1. −−<br />

AE −−<br />

BE<br />

2. D is <strong>the</strong> midpoint of −−<br />

AB .<br />

3. −−−<br />

AD −−<br />

BD<br />

4. −−<br />

ED −−<br />

ED<br />

5. ADE BDE<br />

4 1. −−<br />

TP −−<br />

TQ<br />

2. TPQ TQP<br />

3. SPQ RQP<br />

4. SPT RQT<br />

5. STP RTQ (Vertical angles are<br />

congruent.)<br />

6. STP RTQ<br />

7.<br />

(SAA SAA)<br />

−−<br />

SP −−<br />

PQ<br />

8. −−<br />

PQ −−<br />

PQ<br />

9. SPQ RQP<br />

5 1. −−<br />

EF −−<br />

GF<br />

2. −−<br />

DF −−<br />

HF<br />

3. DFH DFH<br />

4. DFG EFH (SAS SAS)<br />

5. 1 2 (Corresponding<br />

parts of congruent<br />

triangles are<br />

congruent.)<br />

6 1. −−−<br />

AD and −−<br />

FC bisect each o<strong>the</strong>r at G.<br />

2. −−−<br />

AG −−−<br />

DG<br />

3. −−<br />

FG −−<br />

CG<br />

4. AGC DGF (Vertical angles are<br />

congruent.)<br />

5. AGC DGF (SAS SAS)<br />

6. GDE GAC<br />

7. AGB DGE<br />

(Corresponding<br />

parts of congruent<br />

triangles are<br />

congruent.)<br />

8. AGB DGE (ASA ASA)<br />

5-6 Perpendicular<br />

Bisectors of a Line<br />

Segment<br />

(pages 79–80)<br />

1 Given −−<br />

AB and P and Q, such that PA PB<br />

and QA QB, let −−<br />

AB intersect −−<br />

PQ at<br />

point M. APQ BPQ. APM BPM.<br />

APM BPM. −−−<br />

AM −−−<br />

BM . There<strong>for</strong>e,<br />

AMP BMP.<br />

2 a) Given −−<br />

AB and point P such that PA PB;<br />

if two points are each equidistant from <strong>the</strong><br />

endpoints of a line segment, <strong>the</strong>n <strong>the</strong> points<br />

determine <strong>the</strong> perpendicular bisector of <strong>the</strong><br />

line segment. Select a second point that is<br />

equidistant from both A and B, <strong>for</strong> example<br />

<strong>the</strong> midpoint of −−<br />

AB , M. Then −−−<br />

PM is <strong>the</strong> perpendicular<br />

bisector of −−<br />

AB .<br />

b) Given point P on <strong>the</strong> perpendicular bisector<br />

of −−<br />

AB , let point M be <strong>the</strong> midpoint of −−<br />

AB .<br />

PMA PMB.<br />

3 The perpendicular bisector of −−<br />

PQ<br />

Note: Since <strong>the</strong>re are many variations of proofs,<br />

<strong>the</strong> following is simply one set of acceptable<br />

statements to complete each proof. Depending<br />

on <strong>the</strong> textbook used, <strong>the</strong> wording and <strong>for</strong>mat<br />

of reasons may differ, so <strong>the</strong>y have not been<br />

supplied <strong>for</strong> <strong>the</strong> method of congruence applied<br />

in each problem. (These solutions are intended<br />

to be used as a guide—o<strong>the</strong>r possible solutions<br />

may vary.)<br />

4 1. −−<br />

JL −−<br />

KL<br />

2. −−<br />

JM −−−<br />

KM<br />

3. JL KL (Congruent segments are<br />

equal in length.)<br />

4. JM KM<br />

5. −−−<br />

LM is <strong>the</strong> perpendicular bisector of −−<br />

JK .<br />

(Two points, each equidistant from <strong>the</strong><br />

endpoints of a line segment, determine<br />

<strong>the</strong> perpendicular bisector of <strong>the</strong> line<br />

segment.)<br />

6. N is <strong>the</strong> midpoint of JK.<br />

(A point on <strong>the</strong> perpendicular bisector<br />

of a line segment is equidistant from <strong>the</strong><br />

endpoints of <strong>the</strong> line segment.)<br />

7. −−<br />

JN −−−<br />

KN (Definition of a midpoint)<br />

5-6 Perpendicular Bisectors of a Line Segment 23

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