contest-matheuclidean-geometrygeometryplane-geometry
Given $\triangle ABC$ and $\odot(ABC) $, $AG$ and $AD$ are the angle bisectors of $\angle A$ . $CD$ and $AB$ intersects at $E$. $DB$ and $AC$ intersects at $F$. Prove that $E, F, G$ are collinear.
I found that $DC=DB$. By Menelaus theorem I need to prove $$\frac{BG} {CG} \cdot \frac{CF} {AF} \cdot \frac{EA} {EB} =1$$ and do I need to use the angle bisector theorem also?
Let $R$ be the another point (different of $P$) where the $CP$ meets the circumcircle of $APQ$, let $F$ be the point where $BQ$ and $CP$ meets. Since $A$, $P$, $R$ and $Q$ lies in the same circumference it follows $$\measuredangle CRQ =180^{\circ}-\measuredangle PRQ=\measuredangle PAQ=\measuredangle BAC$$ Also, $\measuredangle BAC=\measuredangle PCB=\measuredangle RCB$ for construction, so $\measuredangle CRQ=\measuredangle RCB$ and $QR$ is parallel to $BC$. Then $\angle QRF=\angle FCB=\angle BAC=\angle QBC =\angle FQR$, i.e. $\triangle FBC$ and $\triangle FQR$ are isosceles. It follows that the bisector of $BC$ is perpendicular to $BC$ and passes by $F$, the same for the bisector of $QR$, so the bisector of $BC$ is the bisector of $QR$ too, and it is the line by the circumcircles of $\triangle ABC$ and $\triangle BPQ$
Let the triangle be $ABC$ and external angle bisector of $\angle ABC$ cut $AC$ in $X$, of $\angle ACB$ cut $AB$ in $Y$, of $\angle BAC$ cut $BC$ in $Z$.
By angle bisector theorem, $$\frac{AX}{XC}=-\frac{AB}{BC}... (1)$$ $$\frac{CZ}{ZB}=-\frac{CA}{AB}... (2)$$ $$\frac{BY}{YA}=-\frac{BC}{CA}... (3)$$
(1) ×(2) ×(3) gives, $$\frac{AX.CZ.BY}{XC.ZB.YA}=-1$$ Therefore by converse of Menelaus Theorem X, Y, Z are collinear.
Best Answer
This is a Projective solution. Refer this if you are new to projective. Also the point $H$ in your Diagram, wasn't necessary, so I removed it and introduce a new point called $H$.
Solution: Define $H = DA\cap BC$.
By $lemma$ $2$ from the handout I linked, we get that it is enough to show that $(H,G;C,B)=-1$.
By $lemma$ $5$ from the handout I linked, we get that if we show that $\angle AGH=90$ and $AG$ bisects $\angle BAC$, then we are done. But as $AD, AG$ are angle bisectors we have $\angle HAG =90$ and by question it's given that $AG$ bisects $\angle BAC$. And we are done.