abc is an isosceles triangle in which ab=ac ad bisects exterior angle pac and cd is parallel to ab show - Math - Quadrilaterals

Question

abc is an isosceles triangle in which ab=ac ad bisects exterior
angle pac and cd is parallel to ab show that angle dac=angle bca
and show that abcd is a parallelogram

Vimala · Accepted Answer

Let us observe the following figure

Since ABC is an isosceles triangle we have AB = AC $\Rightarrow ∠ 1 = ∠ 2$ AD is the bisector of external $∠ P A C$ and CD is parallel to AB In a triangle an exterior angle is equal to the sum of two opposite interior angles Thus in triangle ABC, we have $∠ C A P = ∠ 1 + ∠ 2 \Rightarrow ∠ C A P = 2 ∠ 2 [\sin c e ∠ 1 = ∠ 2] \Rightarrow 2 ∠ 3 = 2 ∠ 2 [S i n c e A D i s t h e b i \sec t o r o f e x t ∠ C A P, t h e r e f o r e, ∠ C A P = 2 ∠ 3] \Rightarrow ∠ 3 = ∠ 2 \Rightarrow ∠ D A C = ∠ B C A$ Since AC intersects lines AD and BC at A and C respectively such that $∠ 3 = ∠ 2$ , alternate interior angles are equal Therefore, AD is parallel to BC Given that CD is parallel to AB Thus, ABCD is a parallelogram

abc is an isosceles triangle in which ab=ac.ad bisects exterior angle pac and cd is parallel to ab.show that angle dac=angle bca and show that abcd is a parallelogram