My solution during the exam is the same as the second official solution. Let $D'$ be the reflection of $D$ $w.r.t$ to $A$. Let $F'$ be the perpendicular from $A$ to $BD'$. $Q'$ is the projection of $F'$ onto line $l$. We easily get that $Q'$ is the reflection of $Q$ $w.r.t$ $A$. So it's sufficient to show that $AB \geq PQ'$. Now, notice that AB is the diameter of $\odot AEBF'$. So, $AB \geq EF'$. Also, as $PQ'$ is the projection of $EF'$ onto $l$, $EF' \geq PQ'$. So we get $AB \geq PQ'$.

Is there any other solution Itticantdomath wrote: My solution during the exam is the same as the second official solution. Let $D'$ be the reflection of $D$ $w.r.t$ to $A$. Let $F'$ be the perpendicular from $A$ to $BD'$. $Q'$ is the projection of $F'$ onto line $l$. We easily get that $Q'$ is the reflection of $Q$ $w.r.t$ $A$. So it's sufficient to show that $AB \geq PQ'$. Now, notice that AB is the diameter of $\odot AEBF'$. So, $AB \geq EF'$. Also, as $PQ'$ is the projection of $EF'$ onto $l$, $EF' \geq PQ'$. So we get $AB \geq PQ'$.

Dear Mathlinkers, I don't understand P, Q are the images of E, F on l ? Thank in advance Sincerely Jean-Louis

Projections on $l$.

Let L1 be perpendicular from A to BC. it meets CD at G. Let F' and Q' be reflections of F and Q about L1. we want to prove PQ' ≤ AB. EP and F'Q' are both perpendicular to PQ' so PQ'F'E is trapezoid so PQ' ≤ EF'. now let's prove AB is diameter of AEBF'. we know ∠AF'G = ∠AFG = 90 so we need to prove F',G,B are collinear. ∠F'GA = ∠FGA and ∠CGA = ∠BGA so as C,G,F are collinear we have B,G,F' are collinear as well. so AB is diameter of AEBF' and AB ≥ EF' ≥ PQ' = AP + AQ. we're Done.

$\color{red} \textbf{Geo Marabot Solve 1}$ Take the reflection of the whole figure with respect to the perpendicular bisector of $BC$ and let $X'$ be the reflection of $X.$ So, $$AP+AQ=AP'+AQ=P'Q \le E'F \le AC=AB \blacksquare$$