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.
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.
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!