An acute-angled triangle $ABC$ is given in the plane. The circle with diameter $AB$ intersects altitude $CC'$ and its extension at points $M$ and $N$, and the circle with diameter $AC$ intersects altitude $BB'$ and its extensions at $P$ and $Q$. Prove that the points $M, N, P, Q$ lie on a common circle. (USAMO, 1990)
$MPNQ$ is cyclic if and only if $CBC'B'$ is cyclic. This is because $MPNQ$ is cyclic if and only if $MX\cdot XN=PX\cdot XQ$ by the power of $X$ on the circle whose existence is to be proven. Notice that $B'$ is on the circle with diameter $AB$, because $\angle AB'B=90^{\circ}$, thus $B'X\cdot XB=MX\cdot XN.$
Also, $PX\cdot XQ=CX\cdot XC',$ because $C'$ must be on the circle with diameter $AC$. Therefore, $MX\cdot NX=PX\cdot QX\Longleftrightarrow B'X\cdot XB=CX\cdot XC'$, which happens if and only if $CBC'B'$ is cyclic.
$CBC'B'$ is cyclic. This is because $\angle B'CC'=\angle C'BB'$, since $\triangle ACC'\sim\triangle ABB'.$ Therefore, $MPNQ$ is also cyclic.
I think my solution is likely correct, but my handout gives a different solution, and I'm not very familiar with geometry. Is there anything that I'm missing?
Your solution is correct though it could be made neater.
Let circle with diameter AC be $\omega_C$ and circle with diameter AB be $\omega_B$. As the diameter subtends a right angle on the semicircle, $C'$ lies on $\omega_C$ and $B'$ lies on $\omega_B$.
Now
Power of X wrt circumcircle of $BC'B'C$ is $$CX\cdot C'X = BX\cdot B'X$$
from which follows $$PX\cdot QX = MX\cdot NX$$
Hence proved, $MPNQ$ is cyclic.