# Difficult problem in elementary euclidean geometry

A point D is chosen inside an equilateral triangle $$ABC$$ such that $$AD$$ = $$BD$$. A point $$E$$ outside the triangle is chosen such that $$\angle DBE$$ = $$\angle DBC$$ and $$BE$$ = $$AB$$. Find the degree measure of angle $$\angle DEB$$.

My attempt:

I first tried letting $$\angle EBD$$= x and trying to find the other angles in terms of x, in the hope of getting a congruent triangle. But that didn't go anywhere, or lead to anything useful. It seems that this problem requires some construction to find something equal to $$E$$ but I can't figure that out

• Could you clarify what is angle $E$? Is it $\angle BEA$? – Momo May 28 at 13:44
• @Momo Its angle DEB – MNIShaurya May 28 at 13:51
• Have you tried drawing it ? I think the answer appears quite naturally, making the proof easier – Popyaitte May 28 at 14:19
• @Popyaitte I have. It dosen't seem natural to me, the diagram is really messy-ish. Maybe I'm doing too many unnecessary constructions – MNIShaurya May 28 at 14:38 Note that $$\angle DCB = \frac12\angle C = 30^\circ$$ because of $$AD = BD$$. Since $$\angle EBD = \angle CBD$$, $$EB = AB = BC$$ and $$DB=DB$$, the triangles $$CBD$$ and $$EBD$$ are congruent, which yields $$\angle DEB = \angle DCB = 30^\circ$$.
Construct $$DF\perp BC, E\in BC$$, and $$DG\perp EB, E\in EB$$. Then triangles $$BFD$$ and $$BGD$$ are congruent, and so are $$DGE$$ and $$DFC$$. So $$\angle DEB=\angle DCB = 30^\circ$$. The construction idea comes from the fact that $$D$$ is by construction the incenter of the triangle formed by $$BE$$, $$BC$$, and $$AC$$.
Maybe you can see this as a proof with almost no words. Just think of reflection in the line through $$B$$ and $$D$$: 