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# Geometry - help

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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

Jul 7, 2024

#2
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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.

Jul 7, 2024
#3
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This post fully represents what this site has come to: bot or homework cheater post asinine questions that are answered by other bots or blithering idiots.

Jul 7, 2024
edited by Holtran  Jul 7, 2024