Baltic Way 2010 · Problem 14

We write the power of the point $A$ with respect to the circle $\gamma$ through $D$, $E$, and $C$:

$$|AX||AY| = |AD||AC| = |AC|^2 - |AC||CD|.$$

Similarly, if we calculate the power of $B$ with respect to $\gamma$ we get

$$|BX||BY| = |BC|^2 - |BC||CE|.$$

We have also that $|AC||CD| = |BC||CE|$, the power of the point $C$ with respect to the circle through $A$, $B$, $D$, and $E$. Further if $M$ is the middle point of $XY$ then

$$|AX||AY| = |AM|^2 - |XM|^2 \quad \text{and} \quad |BX||BY| = |BM|^2 - |XM|^2.$$

Combining the four displayed identities we get

$$|AM|^2 - |BM|^2 = |AC|^2 - |BC|^2.$$

By the theorem of Pythagoras the same holds for the point $H$ on $AB$ such that $CH$ is the altitude of the triangle $ABC$. Then since $H$ lies on the side $AB$ we get

$$|AB|(|AM|-|BM|) = |AM|^2 - |BM|^2 = |AC|^2 - |BC|^2 = |AH|^2 - |BH|^2 = |AB|(|AH|-|BH|).$$

We conclude that $M = H$.