Let $EF$ intersect $BC$ at $T$, and let $P$ be the foot of the perpendicular from $D$ onto $EF$. It is known that $(T,D;B,C)=-1$ by the Gergonne concurrency. Since $\angle TPD = 90^{\circ}$, by properties of harmonics we have $PD$ is the bisector of $\angle BPC$, so $\angle MPB=\angle NPC$. Quadrilaterals $MPDB$ and $NPDC$ are cyclic, so this implies $\angle MDB=\angle NDC$. Finally, since $BDC$ is tangent to the incircle this implies that arcs $DP$ and $DQ$ subtend the same angle, so they must be equal, implying $DP=DQ$, as desired.

I’m pretty sure this exact problem has appeared on a past Balkan shortlist or something. I’ll post a link here if I find it. Edit: See JBMO SL 2015/G5.

Alternatively, we want $\angle BDM= \angle CDN$, so proving the similarity of $MBD$ and $CDN$ is sufficient. We want $\frac {MB} {BF}= \frac {CN} {CE}$ which is obvious by LoS in $MBF$ and $CEN$.

ike.chen wrote: I’m pretty sure this exact problem has appeared on a past Balkan shortlist or something. I’ll post a link here if I find it. Yes, it did feel somewhat familiar. I think I saw it in passing, but can't remember exactly where either.

ike.chen wrote: I’m pretty sure this exact problem has appeared on a past Balkan shortlist or something. I’ll post a link here if I find it. https://artofproblemsolving.com/community/c6h1525171p9137012

When i seen this problem on the test i had no doubt about what to do, its just insane how PSC couldnt see this. (Btw this took me 10 mins on test) Let $EF \cap BC=G$, $I$ incenter of $\triangle ABC$, $ID \cap EF=K$ and $D'$ the reflection of $D$ over $I$ ($DD'$ diameter in $\omega$) $$-1=(G, D; B, C) \overset{\infty_{\perp BC}}{=} (G, K; M, N) \overset{D}{=} (D, D'; P, Q)$$Thus we are done

Let $R$ in $EF$ such that $AR$ is parallel to $MB$ and $NC$ It is known that $\frac {MB} {AR}= \frac {BF} {AF}$ and $\frac {AR} {NC}= \frac {AE} {EC}$ . Then we multiply these last 2 expressions and obtain : $\frac {MB} {NC}= \frac {BF} {CE}$ and how $BF$ = $BD$ and $EC$=$DC$ , then $\frac {MB} {NC}= \frac {BD} {CD}$ so that $\triangle MBD$ and $\triangle NCD$ are similar for side-angle-side Then $\angle BDM$ = $\angle CDN$ so that $DP$ = $DQ$ . FINISH

MathLuis wrote: When i seen this problem on the test i had no doubt about what to do, its just insane how PSC couldnt see this. (Btw this took me 10 mins on test) Let $EF \cap BC=G$, $I$ incenter of $\triangle ABC$, $ID \cap EF=K$ and $D'$ the reflection of $D$ over $I$ ($DD'$ diameter in $\omega$) $$-1=(G, D; B, C) \overset{\infty_{\perp BC}}{=} (G, K; M, N) \overset{D}{=} (D, D'; P, Q)$$Thus we are done GOOD SOLUTION , BUT IN THE TEST I COULD NOT NOTICE IT