Two constructed triangles with common vertex have tangent circumcircles at this vertex.

Question

Here's a complicated-sounding geometry problem, Any help would be appreciated :)

Let $\triangle ABC$ be an obtuse-angled triangle with circumcentre $O$, circumcircle $\Gamma$ and $\angle ABC > 90^\circ$. Let $AB$ intersect the line through $C$ perpendicular to $AC$ in $D$. Let $l$ be the line through $D$ perpendicular to $AO$, and let $E$ be the intersection of $l$ and $AC$, and let $F$ be the point between $D$ and $E$ where $l$ intersects $\Gamma$. Can you prove that the circumcircles of triangles $\triangle BFE$ and $\triangle CFD$ are tangent at $F$?

Thanks :D

Answer 1

Hint: The diagram looks somewhat like this. You can solve the problem using simple geometry, I suppose.

Answer 2

Let $l$ intersect $AO$ at $H$, and $AO$ intersect the circle again at $P$. Since $AP$ is a diameter, $PC \perp AC$, so $P, C, D$ are collinear. Now the 2 altitudes $AC, DH$ of $\triangle APD$ intersect at $E$, so $E$ is the orthocenter of $\triangle APD$. Thus $PE \perp AB$. Since $AP$ is a diameter, $PB \perp AB$, so $P, E, B$ are collinear.

$$\angle{PFE}=\angle{PFH}=90^{\circ}-\angle{HPF}=90^{\circ}-\angle{APF}=\angle{PBD}-\angle{DBF}=\angle{PBF}=\angle{EBF}$$

$$\angle{PFC}=\angle{PAC}=90^{\circ}-\angle{APC}=90^{\circ}-\angle{HPD}=\angle{HDP}=\angle{FDC}$$

Thus $PF$ is tangent to both the circumcircle of $\triangle BFE$ and the circumcircle of $\triangle CFD$, so the 2 circumcircles are tangent at $F$.