In the figure, $$ riangle ABC $$ is a right triangle in which,$$ \angle ACB = 90\degree$$,$$ AB=13$$,$$ CD\parallel AB$$, point $$ E$$ is a dynamic point in the ray of $$CD$$ ( not coincide with point $$C$$ ) , connect $$AE$$, intersect $$BC$$ at point $$F$$,$$ \angle BAE $$ bisects $$BC$$ at point $$G$$。

Question

In the figure, $$  	riangle ABC $$ is a right triangle in which,$$ \angle ACB = 90\degree$$,$$ AB=13$$,$$ CD\parallel AB$$, point $$ E$$  is a dynamic point in the ray of $$CD$$  ( not coincide with point $$C$$  ) , connect $$AE$$, intersect $$BC$$ at point $$F$$,$$ \angle BAE $$ bisects $$BC$$ at point $$G$$。

Uniteasy Admin · Accepted Answer

Construct an altitude AH from point $$C$$ to $$AE$$ 
$$S_{	riangle CEF } = \dfrac{1}{2}CH\cdotp EF$$
$$S_{	riangle CAF } = \dfrac{1}{2}CH\cdotp AF  $$
$$S_{	riangle CEF } :  S_{	riangle CAF }  = EF:AF$$
$$\because CE\parallel AB  $$
$$	riangle CFE $$ and $$	riangle BFA$$ are similar 
$$	herefore   EF:AF = CE : AB = \dfrac{3}{13}  $$ $$CE=x$$,$$ AE=y$$ 
$$\because CH \parallel AB $$
$$	herefore  \angle AHC = \angle BAH  $$ 
$$AH $$ biosects $$ \angle BAH$$ 
又$$\angle BAH = \angle EAH   $$
$$	herefore    \angle EAH  =  \angle AHC$$
$$	riangle AEH $$ is an isosceles triangle.
$$AE= HE = y$$
$$CH= CE+EH= x+y$$
又$$	riangle GCH$$ and  $$	riangle GAB$$ are similar 
$$\dfrac{CH}{AB}=\dfrac{CG}{BG} = 2    $$
$$	herefore  \dfrac{x+y}{13}  = 2 $$
$$y = 26-x$$

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