Question

Converse of angle bisector theorem.
Given: In ∆ABC,Dis the point on side BC such that AB upon AC =BD upon DC.
To prove: ray AD bisects angle BAC.
Proof??

Answer 1

Answer:

Step-by-step explanation:

It is given that in ΔABC ray BD bisects  ∠ABC and ray CE bisects ∠ACB and AB=CD.

From ΔABC, BD is the bisector of ∠ABC, then by using angle bisector theorem, we have

              (1)

Also, CE is the angle bisector of ∠ACB, thus

               (2)

Also, we are given that AB=CD, then from (1),(2)and (3)

which implies that segment ED is parallel to segment BC by the converse of basic proportionality theorem.

Hence proved.