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Question

Question 1 (ii)
ΔABC and ΔDBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC (see the given figure). If AD is extended to intersect BC at P, show that
ΔABPΔACP

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Solution

In ΔABD and ΔACD,
AB = AC (given)
BD = CD (given)
AD = AD (common)
ΔABDΔACD (by SSS congruence rule)
BAD=CAD(i)

In ΔABP and ΔACP
AB = AC (given).
BAP = CAP [From (i)]
AP = AP (common)
ΔABPΔACP (by SAS congruence rule)

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