Question

∆ABC and ∆DBC are two isosceles triangles on the same base BC such that AB = AC and DB = DC (AB > BD) and vertices A and D are on the same side of BC. If AD is extended to intersect BC at P, then show that :(a)∆ABD ≅∆ACD(b) APbisects ∠A as well as ∠D (c)∆ABP ≅∆ACP(d) APis the perpendicular bisector of BC

Answer 1

Step-by-step explanation:

1) AB=AC

DB=DC

AD=AD

therefore: tri. ABD=tri.ACD(by S. S. S)

2)AP=AP

AB=AC

ang. C=ang.B(becoz it is an isosceles triangle)

therefore:tri.ABP=tri.ACP(by SAS postulate)

3)tri.ABC is congruent to tri. BDC

p is an angular bisector of bc

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