Chapter 4, Problem 22RP

To determine

To Prove: $AF\perp DE.$

Explanation of Solution

Given information: $AB=AC,BF=FC,\angle BAE=\angle CAD.$

Formula Used: Congruency of triangles.

Proof:

  

  $\begin{array}{l}\angle BAE=\angle CAD\mathrm{....}\left(giving\right)\\ AM=AM\mathrm{....}(common)\\ \therefore \Delta AMB\cong \Delta AMC\mathrm{....}(SAS)\\ \therefore \angle ABD=\angle ACE.\end{array}$

Similarly,

  $\begin{array}{l}\Delta FDM\cong \Delta FEM.\\ \therefore \angle FMD=\angle FME\mathrm{....}(CPCT)\end{array}$

  $\angle FMD+\angle FME={180}^{\circ }$ (Linear pairs)

  $\therefore \angle FMD=\angle FME={90}^{\circ }$