Question

$AD$ is a median of $△ABC$. The bisector of $\angle ADB$ and $\angle ADC$ meet $AB$ and $AC$ in $E$ and $F$ respectively. Prove that $EF||BC$.

Answer 1

In $△DAE$, $DE$ Bisect $\angle ADB$

So we have

$\frac{DA}{DB}=\frac{AE}{EB}$ ------1

Similarly in $△DAC$, $DE$ Bisect $\angle ADC$

we get

$\frac{DA}{DC}=\frac{AF}{FC}$ --- $(DC=DB)$

$\frac{DA}{DB}=\frac{AF}{FC}$ ------2

From 1 and 2

$=\frac{AE}{EB}=\frac{AF}{FC}$

In $△ABC$,

$EF\parallel BE$ (Baisc Proportionality Theorem)