Question
In the given figure, $AB || CD$. Prove that $\angle\text{BAE}-\angle\text{ECD}=\angle\text{AEC}.$
✓
Answer
Draw $\text{EF}||\text{AB}||\text{CD}$ through $E$.
Now, $\text{EF}||\text{AB}$ and $AE$ is the transversal.
Then, $\angle\text{BAE}+\angle\text{AEF}=180^\circ$ [Angles on the same side of a transversal line are supplementary]
Again, $\text{EF}||\text{CD}$ and $CE$ is the transversal.Then,
$\angle\text{DCE}+\angle\text{CEF}=180^\circ$ [Angles on the same side of a transversal line are supplementary]
$\Rightarrow\angle\text{DCE}+(\angle\text{AEC}+\angle\text{AEF})=180^\circ$
$\Rightarrow\angle\text{DCE}+\angle\text{AEC}+180^\circ-\angle\text{BAE}=180^\circ$
$\Rightarrow\angle\text{BAE}-\angle\text{DCE}=\angle\text{AEC}$
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|
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|
|
$150-153$
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