Given a circle $k$ with diameter $AB$, let $C$ be any point on the circle $k$ different from $A$ and $B$. Drop a perpendicular from $C$ to the diameter $AB$, and let the foot of the perpendicular on $AB$ be $D$, and the second intersection point of the perpendicular with the circle $k$ be $E$. The circle centered at $C$ with radius $CD$ intersects the circle $k$ at points $P$ and $Q$. Let the intersection point of segments $CE$ and $PQ$ be $M$. Determine the value of $\frac{PM}{PE} + \frac{QM}{QE}$.
Solution. Using the notation in the figure, $CD = CP = CQ = r$, so by the inscribed angle theorem, $\angle CEP = \angle CEQ = \alpha$. Therefore, in $\triangle PEQ$, $EC$, and thus $EM$ is the angle bisector.

![](https://cdn.mathpix.com/cropped/2024_05_02_f488e6e9282972c3cd6bg-1.jpg?height=786&width=616&top_left_y=298&top_left_x=749)

By the angle bisector theorem:

$$
\frac{PM}{PE} = \frac{QM}{QE}
$$

Using the inscribed angle theorem, $\angle ECQ = \angle EPQ = \beta$, because both are inscribed angles subtending the arc $EQ$ in circle $k$.

Therefore, $\triangle MPE \sim \triangle QCE$, since two of their angles are equal. The reflection of point $C$ over segment $AB$ is point $E$, so $CD = r = DE$, which means $CE = 2r$.

Using these facts:

$$
\frac{QC}{CE} = \frac{1}{2} = \frac{PM}{PE} = \frac{QM}{QE}, \quad \text{so} \quad \frac{PM}{PE} + \frac{QM}{QE} = \frac{1}{2} + \frac{1}{2} = 1
$$

This is the value we were looking for.