ABC is a triangle and the line YCX is parallel to AB such that AX and BY are the angle bisectors of $\displaystyle \angle$A and $\displaystyle \angle$B respectively.If AX meets BC at D and BY meets AC at E and if YE=XD prove that AC=BC.

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- Feb 11th 2011, 05:35 PMabhishekkgpangle bisector problem
ABC is a triangle and the line YCX is parallel to AB such that AX and BY are the angle bisectors of $\displaystyle \angle$A and $\displaystyle \angle$B respectively.If AX meets BC at D and BY meets AC at E and if YE=XD prove that AC=BC.

- Feb 11th 2011, 07:37 PMTKHunny
You've nothing?

Hints:

Alternate Interior Angles of a Transversal cutting two parallel lines

$\displaystyle m\angle X = \frac{1}{2} m\angle A$

$\displaystyle m\angle Y = \frac{1}{2} m\angle B$

$\displaystyle m\angle ACY = m\angle A$

$\displaystyle m\angle BCX = m\angle B$

That's more than nothing and so far I used only one theorem! - Feb 11th 2011, 09:11 PMabhishekkgp
that much i had. also AC=CX and CB=CY. but still i am stuck.