A geometry problem by Lee Dongheng

Let $DF=x$ ,

$EC=AB-116$ , $CG=AE-EC=202+116-AB$

$\triangle ADF \sim \triangle GCF$ ,

Therefore $\frac{DF}{CF}$ = $\frac{AD}{CG}$

$\frac{x}{AB-x}=\frac{AB}{318-AB}$

$318x-ABx=AB^{2}-ABx$

$318x=202^{2}-116^{2}=(202+116)(202-116)$

$\boxed{x=86}$

Bonus question : Can you prove $AE=BE+DF$ ?

$AD = AB = \sqrt{AE ^2 + BE ^2} = \sqrt{202 ^2 + 116 ^2} = \sqrt{27348} = 2\sqrt{6837}$

$\angle BAE = 90 - 2 \angle DAF$

$\angle AEB = 90 - \angle BAE = 90 - (90 - 2 \angle DAF) = 2 \angle DAF$

$\cos \angle AEB = \cos 2\angle DAF= \frac{116}{202} = \frac{58}{101}$

$2\angle DAF = \arccos(\frac{58}{101})$

$\angle DAF = \frac{1}{2} \arccos(\frac{58}{101})$

$\tan\angle DAF= \frac{DF}{2\sqrt{6837}}$

$DF = 2\sqrt{6837}\tan\angle DAF$

$DF = 2\sqrt{6837}\tan(\frac{1}{2} \arccos(\frac{58}{101})) = \boxed{86}$

And;

To prove that $AE = BE + DF$

$\cos (2 \angle DAF) = \frac{BE}{AE} \Rightarrow BE = AE \cos (2\angle DAF) \Rightarrow BE = AE(\cos^2 \angle DAF - \sin^2 \angle DAF)$

$\sin (2 \angle DAF) = \frac{AB}{AE} \Rightarrow AB = AE \sin (2 \angle DAF) \Rightarrow AB = 2 (AE) (\cos \angle DAF)(\sin\angle DAF)$

$tan \angle DAF = \frac{DF}{AD} = \frac{DF}{AB} \Rightarrow AB = \frac{DF}{\tan \angle DAF} \Rightarrow AB = \frac{DF \cos \angle DAF}{\sin \angle DAF}$

$2 (AE) (\cos \angle DAF)(\sin\angle DAF) = \frac{DF \cos \angle DAF}{\sin \angle DAF}$

$DF = 2 (AE) (sin^2 \angle DAF)$

$BE + DF = AE(\cos^2 \angle DAF - \sin^2 \angle DAF) + 2 (AE) (\sin^2 \angle DAF)$

$BE + DF = AE ((\cos^2 \angle DAF - \sin^2 \angle DAF) + 2 (\sin^2 \angle DAF))$

$BE + DF = AE (\cos^2 \angle DAF + \sin^2 \angle DAF)$

$BE + DF = AE (1)$

$\boxed{BE + DF = AE}$