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Parallel Mirage


$  AFG $ is a triangle. $B$ and $C$ are two points on $AF$ and $AG$ such that $FB=BC=CG$. The angle-bisectors of $\angle ABC$ and $\angle ACB$ intersect the sides $AG$ and $AF$ at points $E$ and $D$ respectively. Prove that $DE\mid\mid FG$.


Source: BdMO 2024 National Junior P3


Proof Based Problems  


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Solution

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$BF=BC \implies \angle BCF = \angle BFC$

$Also, $\angle BCF = 180 - \angle BFC - \angle FBC $

$\implies BCF = \frac{180-\angle FBC}{2} = \frac{\angle ABC}{2} = \angle EBC$

So, $BE\mid\mid CF$

In the same way, $BG\mid\mid CD$

\[BE\mid\mid CF \implies \frac{AB}{AF} =\frac{AE}{AC}\]

\[BG\mid\mid CD \implies \frac{AB}{AD} =\frac{AG}{AC}\]

Dividing the equations

\[\frac{AF}{AD} = \frac{AG}{AE}\]

So, $GF\mid\mid DE$

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