# Geometry: Perpendicular tangent

I came up with this but I have not been able to solve it. I would really appreciate any help.

Let $ABC$ be a triangle and let $\omega$ be its circumcircle. Produce the internal angle bisector of $\angle BAC$ to meet $BC$ and $\omega$ in $D$ and $E$, respectively. Let the circle with diameter $DE$ intersect $\omega$ at a second point $F$. Drop a perpendicular from $F$ to "$AC$ produced" and let it meet "$AC$ produced" at $G$. Is it true that line $GF$ is tangent to $\omega$?

(The reason I ask "is it true" is because I don't know if it is.)

• It seems to be false; could you please attach a figure? – user133281 Feb 1 '15 at 1:38
• As pointed out, this is not true. It can also be verified by a moderately accurate drawing using Geogebra. Then, it would be more interesting to ask “If a line from F is drawn to cut AC produced at G, what condition is needed on G such that GF is tangent to ω?” – Mick Feb 1 '15 at 5:41 I was interested in classifying those triangles where your assumption actually holds. Without loss of generality, you can choose a coordinate system in such a way that the circumcircle $\omega$ is the unit circle, and the point $A$ has coordinates $(1,0)$. Then you can describe $B$ and $C$ using a rational representation as $B=(b^2-1,2b)/(b^2+1)$ and $C=(c^2-1,2c)/(c^2+1)$. Then the triangles for which your assumption holds must satisfy
$$b^2c + 2bc^2 + c^3 + 4b = 0$$ 