Question

if the external bisector of the Vertical angle of a triangle is parallel to its base then prove that the triangle is isosceles​

Answer 1

Step-by-step explanation:

Join A to D

Hence,

\begin{array}{l} AD\parallel BC \\ \Rightarrow \angle ACB=\angle CAD\, \, \left( { alternate\, \, angle } \right) \\ \Rightarrow and\, \angle BAC=\angle ACD\left( { alternate\, \, angle } \right) \\ \Rightarrow \end{array}

AD∥BC

⇒∠ACB=∠CAD(alternateangle)

⇒and∠BAC=∠ACD(alternateangle)

⇒

Hence

AB=AC (by ASA congurency )

so, it is isosceles triangle.

solution