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Unformatted text preview: 33. Given: ABC
PQR, BD is an altitude of
ABC. QS is an altitude of PQR.
QP 36. Given: RU bisects
VU RT QS
BD Prove: BA S
Prove: VV
R U V
A CP D R R S QP ABD PQS by AA Similarity and BA
by definition of similar polygons.
34. Given: C
BDA Statements QS
BD SRT. 1. Given
2. Reflexive Prop. 2. SUV STR 3. Corresponding
Postulate 4. AD
BA S 3. B SUV STR 4. AA Similarity S SV
5. VU A Reasons 1. RU bisects
VU RT C
D
C T Proof: Proof: A
P because of the definition of
similar polygons. Since BD and QS are
perpendicular to AC and PR, BDA
QSP. So, AC
Prove: DA SR
RT S Q B SRT. 5. Def. of SR
RT 6. URT VUR 6. Alternate Interior
Theorem 7. VRU URT 7. Def. of 8. VUR VRU 8. Transitive Prop. BDA Given ADB A s ACD AA Similarity A Reflexive Prop. AC
—
DA AD
—
BA Def. of
polygons bisector 9. If 2 of a are
, the sides opp.
these are . 9. VU VR 10. VU VR 10. Def. of S
11. VV
R SR
RT 11. Substitution 35. Given: JF bisects EFG.
EH FG, EF HG
Prove: EK
KF GJ
JF J
E K H F
Proof: RST
ABC, W and D are midpoints of
TS and CB, respectively.
Prove: RWS
ADB 37. Given:
G A Statements Reasons R 1. JF bisects EFG.
EH FG, EF HG 1. Given 2. EFK KFG 3. KFG JKH 4.
5. JKH
EFK EKF
EKF 6. FJH EFK 7.
8. FJH
EKF EKF
GJF 2. Def. of bisector
3. Corresponding
Postulate
4. Vertical are .
5. Transitive Prop.
6. Alternate Interior
Theorem
7. Transitive Prop.
8. AA Similarity 9. EK
KF GJ
JF 9. Def. of B
S D C T W Proof:
RST ABC Given
RS
—
AB s RS
—
AB RS
—
AB polygons 2WS
—
2BD WS
—
BD Def. of Division 178 B
polygons W and D are
midpoints. TS
—
CB Def. of S
Def. of Substitution Chapter 6 segments Given 2WS TS
2BD CB
Def. of midpoint RWS ADB SAS Similarity ...
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This note was uploaded on 12/19/2011 for the course MAC 1140 taught by Professor Dr.zhan during the Fall '10 term at UNF.
 Fall '10
 Dr.Zhan
 Calculus, PreCalculus, Polygons

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