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 $DP = DQ$.

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.