In the figure, \triangle 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

If CE=3, find the ratio of areas of triangles S_{\triangle CEF} : S_{\triangle CAF };

Let CE=x, AE=y. If CG=2GB, find function y in terms of x;

Collected in the board: Triangle

Steven Zheng posted 4 months ago


Construct an altitude AH from point C to AE

S_{\triangle CEF } = \dfrac{1}{2}CH\cdotp EF

S_{\triangle CAF } = \dfrac{1}{2}CH\cdotp AF

S_{\triangle CEF } : S_{\triangle CAF } = EF:AF

\because CE\parallel AB

\triangle CFE and \triangle BFA are similar

\therefore EF:AF = CE : AB = \dfrac{3}{13}

CE=x, AE=y

\because CH \parallel AB

\therefore \angle AHC = \angle BAH

AH biosects \angle BAH

\angle BAH = \angle EAH

\therefore \angle EAH = \angle AHC

\triangle AEH is an isosceles triangle.

AE= HE = y

CH= CE+EH= x+y

\triangle GCH and \triangle GAB are similar

\dfrac{CH}{AB}=\dfrac{CG}{BG} = 2

\therefore \dfrac{x+y}{13} = 2

y = 26-x

Steven Zheng posted 4 months ago

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