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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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