Problem 2

Let

A

be one of the two distinct intersection points of unequal coplanar circles

C_1,C_2

with centers

O_1,O_2

. One common tangent touches them at

P_1,P_2

, and the other at

Q_1,Q_2

. Let

M_1,M_2

be the midpoints of

P_1Q_1,P_2Q_2

. Prove that

\angle O_1AO_2=\angle M_1AM_2

Step 3 of 5: Apply inversion at each center

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\angle O_iAM_i=\angle O_iSA\quad(i=1,2)

Detailed analysis

Invert about the circle $C_i$ , centered at $O_i$ . The relation in Step 2 says that $S$ and $M_i$ are inverse points, while $A$ is fixed because $O_iA$ is the radius. Inversion preserves the angle between the relevant circles/lines, giving $\angle O_iAM_i=\angle O_iSA$ (as directed angles).