Eye-to-Eye Theorem II

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

I shall prove the assertion only for $HIJG.$

Let $E$ be the foot of the perpendicular from $H$ to $AB;$ $K,L,M,N$ points of tangency, as shown in the diagram:$

Triangles $AEH$ and $AMC$ are right and similar because they share an acute angle at $A.$ We have the proportion: $EH/AH=CM/AC$ from which

$\displaystyle EH=R_{C}\frac{R_A}{AC},$

where $R_A$ and $R_C$ are the radii of $(A)$ and $(C),$ respectively. Similarly, for the foot of perpendicular $O$ from $G,$

$\displaystyle GO=R_{C}\frac{R_B}{BC},$

where $R_B$ is the radius of $(B).$

Finally, triangles $ACK$ and $BCL$ are similar due to the choice of $C.$ We have the proportion: $AK/AC=BL/BC,$ i.e., $R_{A}/AC=R_{B}/BC.$ Combining the three proportions shows that $EH=GO.$

Dao Thanh Oai has informed of another solution which is based on a double application of the Eyeball theorem:

One application of the Eyeball theorem shows that $HIQP$ is a rectangle, the other that $SRJG$ is another one. The fact that $PQRS$ is a rectangle due to the central symmetry of diameters $PR$ and $QS$ in circle $(C).$

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.