67th NMO Selection Tests for JBMO

Let $\{T\}=EF\cap BC$. Applying Menelaus' theorem to the triangle $ABC$ and the transversal line $E-F-T$ we obtain: $\frac{TB}{TC}\cdot \frac{EC}{EA}\cdot \frac{FA}{FB}=1$, i.e. $\frac{TB}{TC}\cdot \frac{s-c}{s-a}\cdot \frac{s-a}{s-b}=1$, or $\frac{TB}{TC}=\frac{s-b}{s-c}$, where the notations are the usual ones (1).

This means that triangles $TBN$ and $TCM$ are similar, therefore $\frac{TB}{TC}=\frac{BN}{CM}$. From (1) it follows that $\frac{BN}{CM}=\frac{s-b}{s-c}$, $\frac{BD}{CD}=\frac{s-b}{s-c}$, and $\angle DBN=\angle DCM=90^{\circ }$, which means that triangles $BDN$ and $CDM$ are similar, hence angles $BDN$ and $CDM$ are equal.

This leads to the arcs $DQ$ and $DP$ being equal, and finally to $DP=DQ$.

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Alternative solution.

Let $S$ be the intersection point of the altitude from $A$ with the line $EF$. Lines $BN$, $AS$, $CM$ are parallel, therefore triangles $BNF$ and $ASF$ are similar, as are triangles $ASE$ and $CME$. We obtain $\frac{BN}{AS}=\frac{BF}{FA}$ and $\frac{AS}{CM}=\frac{AE}{EC}$. Multiplying these two together, we obtain $$\frac{BN}{CM}=\frac{BF}{FA}\cdot \frac{AE}{EC}=\frac{BF}{EC}=\frac{BD}{DC}.$$ (We have used that $AE=AF$, $BF=BD$ and $CE=CD$.)

It follows that the right triangles $BDN$ and $CDM$ are similar (SAS), which leads to the same ending as in the first proof.