Let $T\in MN$ such that $BM=BT$ $\Longrightarrow$ $\measuredangle TMB=\measuredangle ETB=\measuredangle FNC$ $\Longrightarrow$ $\triangle TEB\sim \triangle NFC$ $\Longrightarrow$ $\tfrac{MB}{BE}=\tfrac{TB}{BE}=\tfrac{NC}{FC}$, since $BE=BD$ and $FC=CD$ we get $\tfrac{BM}{BD}=\tfrac{CN}{CD}$ then $\triangle DMB\sim \triangle DNC$ $\Longrightarrow$ $\measuredangle MDC=\measuredangle NDC$ $\Longrightarrow$ $\measuredangle PQD=\measuredangle QPD$ hence $DP=DQ$.

If we prove that $\angle PDI= \angle QDI$ we are done, because that implies $\triangle PIB$ and $\triangle QIB$ are congruent. In order to do that, we will prove that $\angle PDB = \angle QDC$. Notice that $\angle AFE= \angle AEF$ and since $MB \parallel NC$ we have $\angle BMF= 180 - \angle ENC$. Hence by law of sines in $\triangle ENC$ and $\triangle BFM$ we get: $\frac{NC}{EC} = \frac{ \sin \angle NEC}{ \sin 180 - \angle BMF} = \frac{ \sin \angle MFB}{ \sin \angle BMF} = \frac{MB}{BF}$ Since $BF=BD$ and $CD=CE$ we get: $\frac{NC}{CD} = \frac{NC}{CE} = \frac{MB}{BF} = \frac{MB}{BD}$ And since $\angle MBD= \angle DCN=90$, that implies: $\triangle MBD \sim \triangle NCD$. Hence $\angle MDB= \angle NDC$ and $\angle PDI= \angle QDI$ Done!

Let's prove $\triangle DMB\sim\triangle DNC$, thus being done. let the $A-$altitude of $\triangle ABC$ intersect $EF$ at $R$. Then $\triangle MBF\sim\triangle RAF$, implying $\frac{MB}{AR}=\frac{BF}{AF}\ (\ 1\ )$ and $\triangle NCE\sim\triangle RAE$, implying $\frac{NC}{AR}=\frac{CE}{AE}\ (\ 2\ )$. Dividing (1) and (2) side by side we get $\frac{MB}{NC}=\frac{BF}{CE}$, but $BF=BD, CE=CD$ hence $\triangle DMB\sim\triangle DNC$, so we are done. Best regards, sunken rock

Let $G$ the intersection of $EF$ and $BC$ ;$H$ the intersection of the perpendicular of $BC$ through $D$ and $EF$ then $(M,N;G,H)=-1$ but $GD \perp DH$ so $DH$ is the angle bisector of $\widehat{PDQ} $ then $\widehat{PDB}=\widehat{QDC}$ therefore $PD=DQ$. R HAS

Hi dude can you send me all jbmo 2016 Tst problems of Romania please?

Suffices prove $DI$ bisects $\angle PDQ$. Let $DI\cap EF=K$ and $EF\cap BC=L$, then we want $(M, N;K, L)=-1$, but $(M, N;K, L)=(B, C;D, L)=-1$, and last equality is obvious.

What is definition for P?.

What is definition for P? GGPiku wrote: Let $ABC$ be an acute triangle with $AB<AC$ and $D,E,F$ be the contact points of the incircle $(I)$ with $BC,AC,AB$. Let $M,N$ be on $EF$ such that $MB \perp BC$ and $NC \perp BC$. $MD$ and $ND$ intersect the $(I)$ in $D$ and $Q$. Prove that $DP=DQ$.

Quote: What is definition for P? Quote: Let $ABC$ be an acute triangle with $AB<AC$ and $D,E,F$ be the contact points of the incircle $(I)$ with $BC,AC,AB$. Let $M,N$ be on $EF$ such that $MB \perp BC$ and $NC \perp BC$. $MD$ and $ND$ intersect the $(I)$ in $D$ and $Q$. Prove that $DP=DQ$. I am quite sure about $P$ is intersection point of incircle and $MD$. There should be typo. My solution: Let $MN \cap BC$ at $X$,$MB \cap ND=T$,$MD \cap NC=Y$ . $(X,D;B,C)=-1$ take pencil from $N$ $\implies$ $(NX,ND;NB,NC)=-1$. $MB$ is parallel to $NC$ so, $MB=BT$ which solves problem.