Given $\triangle ABC$ with $m\angle A = 90^{\circ}$. $D$ is midpoint of $BC$ and $F$ is midpoint of $AB$.

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Given $\triangle ABC$ with $\angle A = 90^{\circ}$. $D$ is midpoint of $BC$ and $F$ is midpoint of $AB$. Points $E$ and $G$ lie on the line $AB$ such that $AE:EF = 1:1$ and $FG:GB = 1:1$ as shown on the picture. What is the ratio of $PQ:PR$?

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What I have done so far.

Using menelaus theorem, I get
$\frac{AE}{EF} \cdot \frac{FC}{CQ} \cdot \frac{QP}{PA} = 1$ or $\frac{PQ}{PA} = \frac{2}{3}$

and $\frac{AE}{EG} \cdot \frac{GC}{CR} \cdot \frac{PR}{PA} = 1$ or $\frac{GC}{CR} \cdot \frac{PR}{PA} = \frac{2}{1}$

I don't have any clue to find $\frac{GC}{CR}$ in order to find $\frac{PQ}{PR}$ after that.

Best Answer

Construct $GX\parallel AD$ ($X\in BC$). We can now use the intercept theorem and see that $AG:GB=DX:XB=4:1$. Let $XB=x$ and so $DX=4x$. We therefore have $DX+XB=5x=DB=CD$ and from here we observe $CD:DX=5:4$. However, $CD:DX=CR:RG$ by the intercept theorem and thus $CG:CR=9:5$.

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