In $\Delta ABC, PQ || BC$ where $P$ and $Q$ are points on $AB$ and $AC$ respectively. The lines $PC$ and $QB$ intersect at $G$. It is also given $EF || BC$, where $G \in EF, E \in AB$ and $F\in AC$ with $PQ = a$ and $EF = b$. Find value of $BC$.
Hi. Hope you are doing well. I was stuck in the geometry problem given above.
I think the Midpoint Theorem has some role to play here but can't figure it out. Please help. This was asked in HOMC 2006. Any help would be appreciated.
On
Draw a line parallel to the side $AB$ through $F$. Such line intersects $BC$ at $K$ and the extension of $PQ$ at $J$. Then the triangles $\triangle ABC$, $\triangle APQ$, $\triangle FJQ$, $\triangle FKC$ are all similar. It follows that $$\frac{b-a}{a}=\frac{c-b}{c}\implies c=\frac{ab}{2a-b}$$ where $c=|BC|$.
$$\frac{EG}{BC}=\frac{PG}{PC}=\frac{1}{1+\frac{GC}{PG}}=\frac{1}{1+\frac{BG}{GQ}}=\frac{GQ}{BQ}=\frac{GF}{BC},$$ which says $EG=FG=\frac{1}{2}b$
Can you end it now?
I got $$BC=\frac{ab}{2a-b}.$$