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The figure below shows parallelogram ABCD and point E is on line BC. Line DE bisects

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The figure below shows parallelogram ABCD and point E is on line BC. Line DE bisects

by Max@Math Revolution » Thu Feb 13, 2020 12:26 am
[GMAT math practice question]

The figure below shows parallelogram ABCD and point E is on line BC. Line DE bisects ∠D. Moreover, BE = DE and ∠A = 120. What is the measure of ∠BDE?
2.13PS.png
A. 20
B. 30
C. 33
D. 40
E. 43
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Re: The figure below shows parallelogram ABCD and point E is on line BC. Line DE bisects

by Max@Math Revolution » Sun Feb 16, 2020 5:49 pm
=>

Since BE = DE, we have ∠DBE = ∠BDE and ∠DEC = 2*(∠BDE) = 2*(∠EDC).
Since ∠EDC + ∠DEC + ∠C = 180°, ∠EDC + 2*(∠EDC) + 120° = 180°, 3*(∠EDC) + 120° = 180°, 3*(∠EDC) = 60°, ∠EDC = 20°.
Then we have ∠BDE = ∠EDC = 20°.

Therefore, the answer is A.
Answer: A
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