Given triangle $ABC$ with right angle $A$ and $D$ is the midpoint of $BC$. $F$ divides $AB$ into two equal parts and $E$ and $G$ divides $AF$ and $FB$ into two equal parts, respectively. The line $AD$ intersects with $CE$, $CF$, and $CG$ at points $P$, $Q$, and $R$, respectively. The ratio of $PQ:QR$ is ...
Here is my solution, using coordinate geometry. I'm looking for a faster way (preferably using just geometry) to solve this problem as it's actually from an old test.
You don't really need the fact that $\angle A = 90^\circ$.
Consider the general case where $Y$ is a point on $AB$ such that $AY:BY = k$:
Using Menelaus' theorem we have
$$\frac{AX}{DX}\cdot \frac{CD}{CB} \cdot \frac{YB}{YA} = 1$$ or $$\frac{AX}{DX} = \frac{CB}{CD}\cdot\frac{YA}{YB} = 2k.$$ So $$\frac{AX}{AD} = \frac{2k}{2k+1}.$$
Thus $$\frac{AR}{AD} = \frac67, \frac{AQ}{AD} = \frac23, \frac{AP}{AD} = \frac27.$$
$PQ:QR$ should come easily from the above.