+297 In parallelogram EFGH, let M be the point on EF such that FM : ME = 3 : 7, and let N be the point on EH such that HN : NE = 2 : 5. Line segments FH and GM intersect at P, and line segments FH and GN intersect at Q. Find PQ / FH.
+9673 Draw the diagram. You will notice that \(\triangle MPF \sim \triangle GPH\) and \(\triangle HNQ \sim \triangle FGQ\).
Using the ratio of corresponding sides in these pairs of similar triangles, we have
\(\dfrac{HQ}{QF} = \dfrac{2}{2 + 5} = \dfrac27\)
and
\(\dfrac{HP}{PF} = \dfrac{3 + 7}7 = \dfrac{10}7\)
We suppose \(HQ : QP : PF = 1 : a : b\) (if not, we just divide through the ratio until the first component is 1).
Then,
\(\begin{cases} \frac1{a + b} = \frac27\\ \frac{1+a}b = \frac{10}7 \end{cases}\)
Cross-multiplying,
\(\begin{cases} 2a + 2b = 7\\ -7a +10b = 7 \end{cases}\)
Solving gives \(a = \dfrac{28}{17}\) and \(b = \dfrac{63}{34}\). Hence, \(HQ : QP : PF = 34 : 56 : 63\).
In particular, \(\dfrac{PQ}{FH} = \dfrac{56}{34 + 56 + 63} = \dfrac{56}{153}\).
