Question

<ABC and <DBC are two isosceles
triangles on the same base BC and vertices A and D are on the same
side of BC. AD is extended to meet BC in P. Show that <ABP=<ACP.​

Answer 1

Step-by-step explanation:

since, ∆ ABC IS AN ISOSCELES TRIANGLE.

THEN,

AB = AC ..........( since two sides of isosceles triangle are equal )

so, angle ABP = angle ACP

because in a triangle angle opposite to equal sides are equal.

Hence, proved.