# Proof of cyclicity using power of a point

Mathematics Asked by David Dong on November 22, 2020

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 $$MXcdot XN=PXcdot XQ$$ 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 $$B’Xcdot XB=MXcdot XN.$$

Also, $$PXcdot XQ=CXcdot XC’,$$ because $$C’$$ must be on the circle with diameter $$AC$$. Therefore, $$MXcdot NX=PXcdot QXLongleftrightarrow B’Xcdot XB=CXcdot XC’$$, which happens if and only if $$CBC’B’$$ is cyclic.

$$CBC’B’$$ is cyclic. This is because $$angle B’CC’=angle C’BB’$$, since $$triangle ACC’simtriangle ABB’.$$ 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?

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 orthocenter $$X$$ wrt $$omega_C$$ is $$CXcdot C'X = PXcdot QX$$
• Power of orthocenter $$X$$ wrt $$omega_B$$ is $$BXcdot B'X = MXcdot NX$$
• Since $$angle BB'C = 90^{circ} = angle BC'C$$ $$BC'B'C$$ is cyclic.

Power of X wrt circumcircle of $$BC'B'C$$ is $$CXcdot C'X = BXcdot B'X$$

from which follows $$PXcdot QX = MXcdot NX$$

Hence proved, $$MPNQ$$ is cyclic.

Answered by cosmo5 on November 22, 2020