∆ ABC and ∆ DBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC . If AD is extended to intersect BC at P , Show that :
i. ∆ ABC = ∆ ACD .
ii. ∆ ABP = ∆ ACP
iii. AP bisects angle A as well as angle D .
iv. AP is the perpendicular bisector of BC .
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Answer:
(I) ∆ ABC=∆ ACD
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