XXVII Olimpiada Matemática Rioplatense

As $T$ is the orthocenter of the triangle $HPQ$, we have that $QT$ is perpendicular to $BE$; then, since $BE$ is perpendicular to $AC$, it follows that $QT$ is parallel to $AC$. Similarly, $TP$ is parallel to $AB$. Assume $QT$ intersects $AB$ at the point $F$ and $TP$ intersects $AC$ at a point $G$.



Since $AP$ is the bisector of the angle $\hat{B}AC$, the triangles $AFQ$ and $AGP$ are isosceles and similar; then: $$\frac{AQ}{AF}=\frac{AP}{AG}(5)$$

On the other hand, as $\hat{A}BP=\hat{A}CQ$, the triangles $ABP$ and $ACQ$ are also similar, so, $$\frac{AQ}{AC}=\frac{AP}{AB}(6)$$

From (5) and (6), we obtain that $\frac{AF}{AG}=\frac{AC}{AB}$. Since the triangles $CAD$ and $BAE$ are similar, then $\frac{AC}{AB}=\frac{AD}{AE}$. Therefore, $$\frac{AF}{AG}=\frac{AD}{AE}.$$

Finally, recalling that $AFTG$ is a parallelogram, we conclude that: $$\text{area}(TDA)=\frac{AD}{AF}\cdot \text{area}(TFA)=\frac{AE}{AG}\cdot \text{area}(TGA)=\text{area}(TAE).$$