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ABCD is a parallelogram. E is midpoint of AB and CE bisects $∠ BCD$ , then $∠ DEC$ is

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60^{o}

90^{o}

100^{o}

120^{o}

answer is B.

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Detailed Solution

Given that, ABCD is parallelogram.
Question Image

E is midpoint of AB and CE bisects

∠

BCD.
We have to find

∠ DEC

.
AB||CD (opposite sides of a parallelogram)
and CE traversal.
We know EC bisects

∠ BCD

in parallelogram ABCD.

∠ D

∠ C = 180

Therefore,

\frac{1}{2} ∠ D + \frac{1}{2} ∠ C = \frac{1}{2} \times 180

\Rightarrow ∠ EDC + ∠ ECD = 90^{0}

△ DEC

∠ DEC + ∠ EDC + ∠ ECD = 180^{0}

∴ ∠ DEC + 90^{0} = 180^{0}

\Rightarrow ∠ DEC = 180^{0} - 90^{0}

\Rightarrow ∠ DEC = 90^{0}

Therefore,

∠ DEC = 90^{0}

.
Hence, option 2 is correct.

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