, 2012

Denote $F$ the orthogonal projection of the point $D$ onto the line $BC$.

Obviously, the points $E$ and $F$ lie on the sides $AC$ and $BC$, respectively. According to Euclid's theorem, $|EC| \cdot |EA| = |CD|^2 - |EC|^2$ and $|FC| \cdot |FB| = |CD|^2 - |FC|^2$,

hence the powers of the point $E$ with respect to the circles $K_1$ and $K_2$ are equal. The same holds for $F$. From this we conclude that $EF$ is the radical axis (or power line) of the circles $K_1$ and $K_2$, hence it passes through the intersection point of these circles, where circles $K_1$ and $K_2$ obviously intersect.
It is sufficient to prove that the angle $\angle ACB$ is a right angle if and only if the point $P$ lies on the line $EF$. If $\angle ACB = \frac{\pi}{2}$, then the quadrilateral $EDFC$ is a rectangle, and $P$ lies on both diagonals of the rectangle because it is the midpoint of one diagonal. Now suppose that $P$ lies on $EF$. According to Thales' theorem, the quadrilateral $EDFC$ is a cyclic quadrilateral with the center of the circumscribed circle in $P$. Using the inscribed angle theorem, we derive $\angle ECF = \frac{1}{2}\angle EPF = \frac{\pi}{2}$. This proves the claim.