Mathematics - Properties of Triangles Question with Solution | TestHub

We have, $△ABC=△ABD+△ACD$ $$\begin{array}{rl}& \Rightarrow \frac{1}{2}bc\sin A=\frac{1}{2}cAD\sin \frac{A}{2}+\frac{1}{2}bAD\sin \frac{A}{2}\\ & \Rightarrow AD=\frac{2bc}{b+c}\cos \frac{A}{2}\end{array}$$ Again, $AE=AD\sec \frac{A}{2}=\frac{2bc}{b+c}$ $\therefore AE$ is $\mathrm{HM}$ of $b$ and $c$. $$\begin{array}{rl}EF& =ED+DF=2DE=2AD\tan \frac{A}{2}\\ & =2\frac{2bc}{b+c}\cos \frac{A}{2}\tan \frac{A}{2}=\frac{4bc}{b+c}\sin \frac{A}{2}\end{array}$$ As $AD\perp EF$ and $DE=DF$ and $AD$ is bisector. $\Rightarrow △AEF$ is isosceles.