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The figure below shows parallelogram ABCD where 2AB = AD.

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by Max@Math Revolution » Wed Feb 12, 2020 12:20 am
[GMAT math practice question]

The figure below shows parallelogram ABCD where 2AB = AD. Moreover, FD = DC = CE and lines BF and AE meet at point P. What is ∠FPE?
2.12PS.png
A. 80
B. 85
C. 90
D. 95
E. 100
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Re: The figure below shows parallelogram ABCD where 2AB = AD.

by Max@Math Revolution » Thu Feb 13, 2020 11:50 pm
=>

Triangles HAB and HDF are congruent, since AB = DF, ∠HAB = ∠HDF and ∠ABH = ∠DFH.
We have AH = DH = (1/2)AD = AB.
Triangles ABG and ECG are congruent using similar reasoning.
Then we have BG = CG = (1/2)AD = AB.
Since we have AH = BG = AB, quadrilateral ABGH is a rhombus and ∠FPE = 90°.

Therefore, C is the answer.
Answer: C
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