Mathematics

# In the above figure, lines $AB$ and $CD$ intersect at $F$. If $\angle EFA=\angle AFD$ and $\angle CFB=50^{0}$, find $\angle EFC$.

##### SOLUTION
In the given figure below,
$AB$ and $CD$ intersect each other at $F$.
$\angle EFA=\angle AFD$ and $\angle CFB=50^{0}$
To find: $\angle EFC$
$\angle AFD=\angle BFC=50^{0}$   (Vertically opposite angles)
But, $\angle EFA=\angle AFD$ (given) = $50^{0}$.
Now, $\angle EFA+\angle EFC+\angle CFB=180^{0}$
(Angles on one side of straight line)
$\Rightarrow 50^{0}+\angle EFC+50^{0}=180^{0}$
$\Rightarrow \angle EFC+100^{0}=180^{0}$
$\Rightarrow \angle EFC=180^{0}-100^{0}=80^{0}$
$\Rightarrow \angle EFC=80^{0}$.

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Subjective Medium Published on 09th 09, 2020
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