# What is the length of $EF$ in the following diagram?

In triangle $$\triangle ABC$$, angle $$A=50^\circ$$ , angle $$C=65^\circ$$ . Point $$F$$ is on $$AC$$ such that, $$BF$$ is perpendicular to $$A$$C. $$D$$ is a point on $$BF$$ (extended) such that $$AD=AB$$. E is a point $$CD$$ such that, AE is perpendicular to $$CD$$. If $$BC=12$$, what is the length of $$EF$$?

I tried and proved that $$ABF \cong AFD$$ and $$BCF \cong CFD$$. I am not able to find any relation of $$EF$$ with other sides.

• $F$ is the midpoint of $BD$ and $E$ is the midpoint of $CD$. Hence $EF={1\over2}BC$. – Aretino Jan 26 at 14:06

From the congruent triangles you have already found, you can conclude that $$\triangle ABC \cong \triangle ACD.$$ Use the two known angles of $$\triangle ABC$$ to find the third angle. You will then be able to show that $$\triangle ABC$$ and $$\triangle ACD$$ are isosceles triangles, and that $$E$$ is the midpoint of $$CD.$$
Since $$AB = AD,$$ triangle $$\triangle ABD$$ also is isosceles and $$F$$ is the midpoint of $$BD.$$
These facts should tell you something about the relationship between $$\triangle BCD$$ and $$\triangle FED,$$ after which you can find the answer.
Angle $$CBA=180°-50°-65°=65°=\angle ACB$$, so $$AB=AC=AD$$. In triangle $$BCF$$, $$CE=BC/2$$, $$FC=BC\cos 65°$$ and $$\angle FCE = 65°$$, so one can find $$FE$$ with cosine theorem.