Question

In the figure, prove that ΔABP ≅ ΔACP, Also prove BP = PC

Answer 1

Answer:

Step-by-step explanation:

since AB = AC , It's an isosceles triangle.

∴AB = AC                                      (given)

 ∠ABP = ∠ACP                             ( property of isosceles triangle)

AP=AP                                           (common)

∴ΔABP ≅ ΔACP

∴BP=PC                                      (∵ Congruent parts of congruent triangles)

Answer 2

Answer:

Angle ABP is perpendicular with angle ACP

So,90+90=180

proved ...

Hope it helps to u