Internal bisector of A of A B C meets side B C at D . A line drawn...

We have, $\triangle A B C=\triangle A B D+\triangle A C D$$\begin{aligned}& \Rightarrow \frac{1}{2}b c \sin A=\frac{1}{2}c A D \sin \frac{A}{2}+\frac{1}{2}b A D \sin \frac{A}{2}\\ & \Rightarrow \quad A D=\frac{2 b c}{b+c} \cos \frac{A}{2} \end{aligned}$Again, $\quad A E=A D \sec \frac{A}{2}=\frac{2 b c}{b+c}\therefore A E$ is $\mathrm{HM}$ of b and c.\begin{aligned}E F & =E D+D F=2 D E=2 A D \tan \frac{A}{2}\\ & =2 \frac{2 b c}{b+c} \cos \frac{A}{2} \tan \frac{A}{2}=\frac{4 b c}{b+c} \sin \frac{A}{2} \end{aligned}$$As$A D \perp E F$and D E=D F and A D is bisector.$\Rightarrow \triangle A E F$ is isosceles.