# Euclid Geometry: Seeking for a simpler geometric solution

The problem:

Extend $$AC$$ to the side of $$C$$, such that $$BC=CE$$. Then, construct the equilateral triangle $$BEF$$. Triangles $$CAB$$ and $$EFC$$ are congruent, thus $$AC=BE$$. Now, observe that $$AC=DE$$, thus $$BE=DE$$. It follows that $$\triangle BED$$ is isosceles, thus $$\angle EDB=80$$.

I was given this problem. The provided construction is smart, but I have a feeling that there should be a simpler non-trigonometric approach. Can anyone think of one?

• Well, I have a solution which also use a constructon of an equilateral triangle. I will post a solution tommorow if you are interested. – Maria Mazur Apr 6 at 22:12
• @Maria Sure. Please do. – blackened Apr 7 at 2:14

Say $$O$$ is a circumcenter of circumcircle $$(ABC)$$ and let $$BO$$ cut $$AC$$ at $$E$$. Then $$\angle COB = 2 \angle CAB = 60$$ so triangle $$BCO$$ is equaliteral. So we have $$BC =OC=OB = OA$$ Since $$\angle AOB = 2 \angle ACB = 80$$ and since $$\angle OAE = 20$$ (note that $$ACO$$ is isosceles) we have $$\angle AEO = 80$$ so $$AO=AE$$ which means $$E =D$$, and finaly we have $$\angle CDB =80$$.
• Thanks. This is also a clever solution. But would you assert that it is simple? One first has to forget about point $D$ (and thus the fact that $AC=BE$), do some auxiliary constructions, then show that ${E}={D}$. – blackened Apr 7 at 8:18