Geometry || greprepclub question

GRE Quant Problem Solving
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Question
Question819×586 29.2 KB

https://greprepclub.com/forum/ab-de-bc-cd-be-is-parallel-to-cd-and-bc-is-parallel-to-df-5831.html

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I edited the source for you this time but please mention the source of question from next time.

AB = AE = lets say 2
Now join BD and CF, so, point where CF meet on BD = X
You will get 4 rt tringles in Quad. BCDF

Area of ABE = \frac{b\times h}{2}=\frac{4}{2}=2
Area of one of the rt. triangles of Quad. BCDF = \frac{b\times h}{2}=\frac{1}{2} and we got 4 of them so, 4 \times \frac{1}{2}=2
Ans C

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IMG_20210826_224232__01
IMG_20210826_224232__011800×2730 1.19 MB

Neglect the below diagram I’m bad at drawing :joy::joy:

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Thanks for adding the source of the question. I will take care of it going forward.

Could you please explain how AB = AE?

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Side opposite to equal angle are equal
image