Question
In figure AE bisects $\angle\text{CAD}$ and $\angle\text{B}=\angle\text{C}.$ Prove that AE || BC.
✓
Answer
Let $\angle\text{B}=\angle\text{C}=\text{x}$ Then,$\angle\text{CAD}=\angle\text{B}+\angle\text{C}=2\text{x}$ (exterior angle)
$\Rightarrow\frac{1}{2}\angle\text{CAD}=\text{x}$
$\Rightarrow\angle\text{EAC}=\text{x}$
$\Rightarrow\angle\text{EAC}=\angle\text{C}$
These are alternate interior angles for the lines AE and BC$\therefore\text{AE }||\text{ BC}$
$\Rightarrow\frac{1}{2}\angle\text{CAD}=\text{x}$
$\Rightarrow\angle\text{EAC}=\text{x}$
$\Rightarrow\angle\text{EAC}=\angle\text{C}$
These are alternate interior angles for the lines AE and BC$\therefore\text{AE }||\text{ BC}$
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