One liner

Ok not as elegant as above but, let \(P_1, P_2\) the feet of \(\perp\) from \(B, F\) on \(AC, AE\), clearly concurrent on \(AD\) by Sondat's theorem. Also, by PoP \(AP_1 \times AC = AP_2 \times AE = AB^ 2\), implying the conclusion.

This problem was proposed by Burii.

2019 Baltic Way P13 wrote: Let $ABCDEF$ be a convex hexagon in which $AB=AF$, $BC=CD$, $DE=EF$ and $\angle ABC = \angle EFA = 90^{\circ}$. Prove that $AD\perp CE$.

Bashy One-Liner

let $w_1,w_2$ two circles with Centers $C,E$ and radius circle $BC,EF$. $AD\perp EC$ if we show $A,D$ are in common chord of $w_1,w_2$ point $D$ lie on two circles and $AB,AF$ are tangents from $A$ to $w_1,w_2$ and $AB=AF$. so $A$ lie on common chord of $w_1,w_2$ so $AD\perp EC$.

Attachments:

A little more fleshed out. Let $\omega_1$ and $\omega_2$ denote the circumcircles of $FED$ and $DCB$ respectively. I claim that $A$ lies on the radical axis of $\omega_1$ and $\omega_2$. This is equivalent to the statement $pow(A,\omega_1) = pow(A,\omega_2)$ which is obviously true because $AE^2-FE^2=AC^2-BC^2$ by the original condition. Therefore $AD$ is the radical axis of $\omega_1$ and $\omega_2$, which implies the result.