+118 In parallelogram EFGH let M be the point on EF such that FM:ME = 1:2, and let N be the point on EH such that HN:NE = 1:3. Line segments FH and GM intersect at P and line segments FH and GN intersect at Q. Find PQ/FH
+1839 Now, we want to find the ratio PQ/FH. We can find the lengths of PQ and FH by using similar triangles. Let's look at triangle HFG and triangle GNF.
Since HM is parallel to GN, we have angle NFG = angle GFM. Additionally, since FM/MG = 1/2, we have FG = 2GF. Similarly, using triangle HEN and triangle EMF, we find that EH = 3HE.
Now we see that FG:GF = EH:HE = 3:2. Therefore, triangles HFG and GNF are similar.
From here, we know that PQ/FH = QF/FH = HN/HF. By letting HF = a and HN = b, we have a+b = FH. Since HN:NE = 1:3, b/3y = 1/3, b = y. Similarly, we know that a/x = 2x/x = 2, a = 2x.
Now we have PQ/FH = HN/HF = y/(a+b) = y/(2x+y). Let Y = PQ/FH = y/(2x+y). Since y = HN = 1 and x = FM = 1, we have Y = 1/(2+1) = 1/3.
Therefore, the ratio PQ/FH is 1/3.
