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# help me help

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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.

Apr 30, 2024

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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, $$\dfrac{PQ}{FH} = \dfrac{56}{34 + 56 + 63} = \dfrac{56}{153}$$.

Apr 30, 2024