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Question
1. In the Fig. 11.50
that (1) AABD = AACT
In the Fig. 11.51, ABC
of ZBAC intersect BCA
3. In the Fig. 11.52
11 50. ABC is a triangle in which AB = AC and D is the mid-point of BC. Prove
AABD = AACD, (ii) ZADB = 90°. If ZBAD = 37°, show that LACD = 530
11 51. ABC is a triangle right angled at B such that AB = BC. The internal bisector
Cintersect BC at D. Prove that AC = AB + BD.
in 11.52, line segment PQ and RS intersect at O. PR = PS and QR = QS. Prove that
GOR = OS, (ii) Z POS = 90°.
R
> 675 1125
-
SD
Fig. 11.50
BD
Fig. 11.51
Fig. 11.52
int of the side BC of a AABC. PQ 1 AB and PR I AC
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