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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.

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

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  • $\begingroup$ Thank you for the answer. I successfully found that EF = 6. That was a great explanation! $\endgroup$ – Shromi Jan 26 at 15:10
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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.

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