# Maximum Perimeter of triangle in a Square

In square $ABCD$, the length of its sides is $5$.

$E$, $F$ are two points on $AB$ and $AD$ in such a way so that $\angle ECF = 45^{\circ}$.

Find the maximum value of the perimeter of $\Delta AEF$.

Let say the $\angle ECB=\theta$ then $\angle DCF=45^{\circ}-\theta$. Now note that $AE=5(1-\tan \theta )$, $AF=5(1-\tan(45^{\circ}-\theta))$ and $EF=\sqrt{AE^2+AF^2}=5(\tan\theta + \tan(45^{\circ}-\theta))$ by elementary trig, the simplification for EF goes like this
So perimeter of trianlge $AEF$ is always of $10$ now matter where $E$ or $F$ are on $AB$ and $AD$.
• $AE+AF+EF≠10$. EF is wrong. – Takahiro Waki Jan 8 '17 at 6:05