Question

In△ABC, AB = AC. If the interior circle of△ABC touches the sides AB, BC and CA at D, E, F respectively. Prove that E bisects BC.

*No spam please*

Answer 1

AB = AC... given 1
AD = AF.... Tangents from exterior point to circle are equal 2
Similarly, BD = BE 3
CF = CE 4
Subracting 2 from 1
DB = FC 5
From 3, 4 and 5
BE = CE
therefore, E bisected BC

Answer 2

Answer:

$\ E bisects B C .$

Step-by-step explanation:

$A F=A D\\ B E=B D \text {, }$  (Tangents from external points)

$\begin{aligned}C E &=C E \\A B &=A C \\A D+B D &=A F+F C \\B D &=F C \quad(\because A D=A F) \\B E &=E C(\because B D=B E, C E=C F)\end{aligned}$

$\therefore E bisects B C .$