Raja Oktovin wrote: Let $ ABC$ be a triangle. A circle $ P$ is internally tangent to the circumcircle of triangle $ ABC$ at $ A$ and tangent to $ BC$ at $ D$. Let $ AD$ meets the circumcircle of $ ABC$ agin at $ Q$. Let $ O$ be the circumcenter of triangle $ ABC$. If the line $ AO$ bisects $ \angle DAC$, prove that the circle centered at $ Q$ passing through $ B$, circle $ P$, and the perpendicular line of $ AD$ from $ B$, are all concurrent. Let E be the intersection of (Q) and AC, O is the circumcenter of triangle ABC We have $ \frac{AP}{AO}=\frac{AD}{AQ}$ then $ PD//OQ$ or $ OQ\perp BC$ So Q is the midpoint of arc BC Since PO bisects $ \angle DAC$, it's easy to see triangle AQC and ADE are isosceles then DECQ is a isosceles trapezium. We get $ \angle QED=\angle EDC=\angle DAE$ So QE is a tangent of (P) Thus $ QE^2=QD.QA=QB^2$ We obtain $ QB=QE$ or E lies on (Q) Moreover, $ \angle DQE=\angle DCE=\angle AQB$ then $ BQE$ is an isosceles triangle Therefore $ AD\perp BE$ Hence the circle centered at $ Q$ passing through $ B$, circle $ P$, and the perpendicular line of $ AD$ from $ B$, are all concurrent at E