A geometry problem by Suvajyoti Barman
Geometry
Level
3
In the given figure,BD=CD,BE=DE,AP=PD DG||CF. Find area of ADH/area of ABC
1/3
1/5
1/4
none of the rest
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Looking purely at the lengths of the bottom segments, AD cuts the entire triangle in half and AE cuts one of those halves in half. This makes makes triangle AED 1/4th the original triangle, but we're looking for ADH, which is clearly less than that amount. This at least rules out 1/4 and 1/3 as answers.
I'm not sure how to reach 1/5 as the correct answer, but for that to be the case, triangle HED has to be 1/20th of the original full triangle since 1/5 + 1/20 = 1/4.
My reasoning may be off but there may be another way to reach 1/5 if it can be determined that the triangle formed by A, P and the unlabeled point they touch is 1/20 because ADH is a similar triangle that is 4 times as large (it's side lengths are doubled).
We also have a lot of information about the side lengths of ADH. HE is the only length we don't yet have a relationship established with some other part. There may be a way to calculate the area as a proportion of the area of AED which we know is 1/4.