Quit easily I've been able to show that triangle AED is similar to triangle BEC by angle-angle (we're given the two angles, and AED and BEC are vertical angles).
I'm absolutely stuck with where to go next.
I want to show it using similar triangles.
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There is no angle P in the picture, that should be E? – imranfat Oct 02 '15 at 00:36
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Yes, I guess it never change, I'll go back and edit it. – Charles Oct 02 '15 at 00:54
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There is a way to show this using circles, let me know if you are interested and I will post the proof. – alexgiorev Dec 10 '15 at 09:01
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You have $\Delta ADE \sim \Delta BCE$, then $\frac{AE}{BE} = \frac{DE}{CE}$, so $\frac{AE}{DE} = \frac{BE}{CE}$. We get $\Delta ABE = \Delta DCE$. Then $\angle ABE = \angle ECD$.
GAVD
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