Two Diameters and Longest Common Chord

Circles $(A)$ and $(B)$ intersect in two points $M$ and $P$. A common chord $DE$ is drawn through $M$ with $D$ on $(B)$ and $E$ on $(A)$.

Prove $EP$ is a diameter of $(A)$ if and only if $DP$ is a diameter of $(B).$

(The applet below illustrates the problem.)

Created with GeoGebra

Solution

Circles $(A)$ and $(B)$ intersect in two points $M$ and $P$. A common chord $DE$ is drawn through $M$ with $D$ on $(B)$ and $E$ on $(A)$.

Prove $EP$ is a diameter of $(A)$ if and only if $DP$ is a diameter of $(B).$

Solution

Note that the angles in $\Delta DEP$ do not depend on the position of point $D$, for angles at $D$ and $E$ are inscribed and always subtended by the same chords. Among all triangles $DEP$ the largest has the largest possible side $DP,$ which is exactly the diameter of $(B).$ Applying this same argument in $(A)$ shows that for the largest $\Delta DEP,$ $EP$ is a diameter of $(A).$

This is a redux of the problem that required to find the longest common chord ($D'F'$ in this case) of two intersecting circles.