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In the figure if $BD ∥ EF$ , then $∠ CEF$ is

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100^{o}

120^{o}

140^{o}

160^{o}

answer is C.

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Detailed Solution

The given figure is,
Question Image

Now, on the line BD, we have,
A right angle at B and a point C on line BD where it forms two rays AC and CE.
A line EF originates at E such that EF is parallel to BD.

BD | | EF

and

∠ ECD = 40^{0}

.
As

BD | | EF

and line

CE

join both lines then we have,

∠ CEF + ∠ ECD = 180^{0}

(As interior angles are supplementary)

\Rightarrow ∠ CEF = 180 ° - 40 °

\Rightarrow ∠ CEF

= 140^{0}

.
Hence,

∠ CEF is 140 °

.
Therefore, option 3 is the correct answer.

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