48th Austrian Mathematical Olympiad

See Figure 3. As usual, let $\beta =\angle ABC$ and $\gamma =\angle ACB$. Since we know that

Figure 3: Problem 13

$\angle AFH=\angle AEH=90^{\circ }$ holds, the quadrilateral $AFHE$ is cyclic, and because $DA$ is parallel to $PQ$ we obtain $$\angle FQP=\angle FAH=\angle FEH=\angle FEP.$$ It follows that $QFPE$ is also cyclic. Since $\angle AFC=\angle ADC=90^{\circ }$, $AFDC$ is also cyclic, and we have $\angle QFP=\angle AFD=180^{\circ }-\angle ACD=180^{\circ }-\gamma$. We therefore have $\angle QEP=\gamma$. From this, we obtain $\angle EAN=90^{\circ }-\gamma =\angle AEP-\angle QEP=\angle AEN$, which shows us that triangle $ANE$ is isosceles. It therefore follows that $N$ is the circumcenter of the right triangle $AHE$, and we therefore have $NA=NH$, as claimed.