congruences-geometrycontest-mathgeometrytriangles
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
$EF$?
Source: Bangladesh Math Olympiad 2016 Junior Catagory
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.
Let $E$ be on $BC$ so $CE = CD$. Then $\angle DEC = 80^{\circ}$ so $CADE$ is cyclic, so we have by the power of the point with respect to $B$: $$a\cdot (a-d) = b\cdot (b-x)\;\;\;\;\;\;\;\;\;\;\;\;(1)$$
By angle bisector theorem we have $${b-x\over x} = {a\over b}\implies b(b-x)= ax\;\;\;\;\;\;\;\;(2)$$
By (1) and (2) we have $$a(a-d)= ax \implies a-d= x$$
Since we are searching for $d+x = a= 2017$ we are done.
$\triangle ABC \sim \triangle AMM' \sim \triangle PNM'$
$\frac{BC}{AB}=\frac{NM'}{PN}$
$PM'=PM=2PN$
$NM'=\sqrt{(2PN)^2-PN^2}=PN\sqrt3$
$\frac{NM'}{PN}=\sqrt3$
$\frac{BC}{AB}=\sqrt3$
$BC=AB\sqrt3=5\sqrt3$
$a=5,b=3,\color{blue}{a+b=8}$
Best Answer
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.