Garcia's Two Circles Lemma
What is this about?
Problem
In $\Delta ABC,$ a circle with $BC$ as a chord meets $AC$ at $G$ and $AB$ at $E.$ Another circle with $AC$ as a chord meets $BC$ at $K$ and $AB$ at $M.$ $N$ is the intersection of $KM$ and $EG.$
Prove that $EN=MN.$
Proof
The proof employs angle chasing. Let $\angle ACB=\gamma.$ Since quadrilateral $AMKC$ is cyclic, $\angle ACK+\angle AMK=180^{\circ},$ implying $\angle AMN=180^{\circ}-\gamma.$ In other words, $\angle EMN=\gamma.$ Similarly, since quadrilateral $BEGC$ is cyclic, $\angle MEN=\gamma.$
It follows that $\Delta EMN$ is isosceles and $EN=MN,$ as required.
Note: The circles may cross the side lines of the triangle in points either interior or exterior to the sides. The reasoning in cases other than that considered above requires only minor adjustments.
Acknowledgment
The statement and the proof are due to Emmanuel Antonio José García (Dominican Republic).
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