Problem Set B, continued 24 Given: PÝ bisects ZVPZ. (2x + 7)°, LVPY ZZPY = (3x - 9)°, PZ = ¹/x + 5, PV = x - 3 Prove: AVPY= AZPY (Use a paragraph proof.) 25 Given: 23 = 41, 44 = 42, ZDAC = 23, ZBAC = 21, AD AB Prove: ACAD= ACAB Problem Set C 26 Given: AB = AE; Z A D P 3 4 (1-x2107) A 1/2 E 2x+7 D

Algebra: Structure And Method, Book 1
(REV)00th Edition
ISBN:9780395977224
Author:Richard G. Brown, Mary P. Dolciani, Robert H. Sorgenfrey, William L. Cole
Publisher:Richard G. Brown, Mary P. Dolciani, Robert H. Sorgenfrey, William L. Cole
Chapter4: Polynomials
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How to do proof #25

Problem Set B, continued
24 Given: PỶ bisects ZVPZ.
LVPY= (2x + 7)º,
ZZPY =
-
(3x − 9)º,
PZ = 1/x + 5,
PV = x - 3
Prove: AVPY= AZPY
(Use a paragraph proof.)
25 Given: 23 = 41, 44 = 42,
ZDAC = 23, ZBAC = 21,
AD = AB
Prove: ACAD= ACAB
Problem Set C
26 Given: AB = AE;
AE and AC trisect ZBAD.
AB 1 BC,
AE 1 DE
Conclusion: AABC = AAED
Z
D
一人のブ
A
17
+2x+2
3 4
B
Y
N
E
V
Jun
E
>C
D
Transcribed Image Text:Problem Set B, continued 24 Given: PỶ bisects ZVPZ. LVPY= (2x + 7)º, ZZPY = - (3x − 9)º, PZ = 1/x + 5, PV = x - 3 Prove: AVPY= AZPY (Use a paragraph proof.) 25 Given: 23 = 41, 44 = 42, ZDAC = 23, ZBAC = 21, AD = AB Prove: ACAD= ACAB Problem Set C 26 Given: AB = AE; AE and AC trisect ZBAD. AB 1 BC, AE 1 DE Conclusion: AABC = AAED Z D 一人のブ A 17 +2x+2 3 4 B Y N E V Jun E >C D
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