1. AABC and A DBC are two isosceles triangles on
the same base BC and vertices A and D are on the
same side of BC (see Fig. 7.39). IfAD is extended
to intersect BC at P, show that
(1) ΔΑΒDeΔΑCD
B
(ii) A.ABP=AACP
(iii) AP bisects Z A as well as Z D.
(iv) AP is the perpendicular bisector of BC.
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