# Eye-to-Eye Theorem II

### Problem

$(A)$ and $(B)$ are two non-intersecting circles, with the internal center of similarity at $C.$ $(C)$ any circle with the radius less than $AC$ and $BC.$ From $A$ and $B$ draw tangents to $(C)$ and mark points $M,N,I,H$ and $G,J,P,O$ where they intersect circles $(A)$ and $(B),$ respectively.

Then points $M,H,G,O$ are collinear as are points $N,I,J,P;$ both lines are parallel to $AB,$ making several rectangles, say, $HIJG$ and $MNPO.$

### Hint

There is a good deal of tangent lines and, hence, right triangles. Some of these are similar.

### Solution 1

I shall prove the assertion only for $HIJG.$

### Acknowledgment

The problem arouse as a generalization of the one I dubbed Eye-to-Eye Theorem I and is entitled to the moniker of Eye-to-Eye Theorem II.

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