In parallelogram $ABCD$, angle $A$ is acute, point $E$ is on the $AD$ such that $BE$ is perpendicular to $AD$ and point $F$ is on line $CD$ such that $BF$ is perpendicular to $CD$. If $AB=BF=13$ and $AE=5$, compute the length of $EF$.
I drew the diagram, all I can do is to get the length of $BE$ by Pythagorean theorem, I'm learning geometry (sophomore), does this question require knowledge about calculus or trig? Can anyone help me with it?
This question is from senior A division contest.
There's another method than what Blue said in the comments (sorry!) but involves knowledge of basic trigonometry.
By the Pythagorean Theorem, we have $BE=12$. We can find $\angle BAE$ using any of the trigonometric ratios (I'll use $\tan$): $$\tan\angle BAE=\frac{12}{5}\Rightarrow \angle BAE \approx 67.38^\circ$$ Now, since $ABCD$ is a parallelogram, we have $\angle BCF=67.38^\circ$ also.
Next, we have to find $\angle EBF$. We know that
$$\angle EBF = \angle ABC - (\angle ABE+\angle CBF)=\angle ABC - 2\angle ABE$$
$\angle ABC=180-\angle BAE=112.62^\circ$ since adjacent angles in a parallelogram must add up to $180^\circ$. $\angle ABE = 90 - \angle BAE=22.62^\circ$ since the acute angles in a right triangle must sum to $90^\circ$. Hence, we have
$$\angle EBF=112.62-2(22.62)=67.38^\circ$$
Finally, we can use the cosine law to find $EF$. We have:
$$EF=\sqrt{12^2+13^2-2\times 12\times 13\times \cos67.38}$$ $$=\sqrt{193}$$