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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