Yet Another Seven Circles Theorem
What Might This Be About?
Let $P$ be a point in the plane of $\Delta ABC.$ $AP,$ $BP,$ $CP$ meet the circumcircle $(O)$ of $\Delta ABC$ again at $A_1,$ $B_1,$ $C_1,$ respectively. Define $A_b$ and $A_c$ as the circumcenters of $\Delta APB_1$ and $\Delta APC_1.$ Define $B_a,$ $B_c,$ $C_a,$ $C_b$ cyclically.
Prove that $A_bB_a,$ $B_cC_b,$ and $A_cC_a$ are concurrent at a point, say $D,$ half way between $P$ and $O.$
The solution is based on the Isosceles Trapezoid lemma. As seen on the diagram and has been shown in the proof of the lemma, the quadrilateral $OB_cPC_b$ is a parallelogram:
It follows that $B_cC_b$ and $OP$ intersect at their midpoints. The same holds for $A_bB_a$ and $A_cC_a$ and, so all four segments concur at a their shared midpoint.