Maximum Perimeter of triangle in a Square

Question

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 $\Delta AEF$.

Answer 1

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
\begin{align*} EF^2&=AE^2+AF^2=5^2((1-\tan \theta)^2)+(1-\tan (45^{\circ}-\theta))^2)\\ &=5^2(2+\tan^2 \theta + \tan^2 (45^{\circ}-\theta) -2(\tan\theta+\tan(45^{\circ}-\theta)))\\ &=5^2(2+\tan^2 \theta + \tan^2 (45^{\circ}-\theta) -2\tan45^{\circ}(1-\tan \theta \tan(45^{\circ}-\theta) )) ,\text{using}\tan(A+B)=\frac{\tan A+\tan B}{1-\tan A \tan B}\\ &=5^2(\tan \theta + \tan(45^{\circ} - \theta))^2 \end{align*}

So perimeter of trianlge $AEF$ is always of $10$ now matter where $E$ or $F$ are on $AB$ and $AD$.