Given a convex quadrilateral, assume that angle ADB = angle ACB. show that angle ABD= angle ACD

Question

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.

Answer 1

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 $\angle ABE = \angle ECD$.