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ADBA – ADAC [by SAS-similarity]
ZB = Z2 and Z1 = ZC
Z1 + Z2 = ZB + ZC ZA= ZB + ZC
2 ZA= ZA+ ZB + ZC = 180°
→ ZA= ZBAC = 90°.
EXAMPI
P
17 In the given figure PA, QB and RC
each is perpendicular to AC such that
PA = x, RC = y, QB = Z, AB = a and
BC = b.
Prove that
11_1
[CBSE 2000C]
SOLU
Х
+
N
х
y
Z
A
а
B b c
PAL AC and QB I ACQB || PA.
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