Question
In the given figure, $BA || ED$ and $BC || EF$. Show that $\angle\text{ABC}+\angle\text{DEF}=180^\circ.$
✓
Answer
Construction: Produce $ED$ to meet $BC$ at $Z$.
Now, $AB || EZ$ and $BC$ is the transversal.
$\Rightarrow\angle\text{ABZ}+\angle\text{EZB}=180^\circ$ (interior angles)
$\Rightarrow\angle\text{ABC}+\angle\text{EZB}=180^\circ\ .....(\text{i)}$
Also, $EF || BC$ and $EZ$ is the transversal.
$\Rightarrow\angle\text{BZE}=\angle\text{ZEF}$ (alternate angles)
$\Rightarrow\text{BZE}=\angle\text{DEF}\ .....(\text{ii)}$ From $(i)$ and $(ii)$,
we have $\angle\text{ABC}+\angle\text{DEF}=180^\circ$
Now, $AB || EZ$ and $BC$ is the transversal.
$\Rightarrow\angle\text{ABZ}+\angle\text{EZB}=180^\circ$ (interior angles)
$\Rightarrow\angle\text{ABC}+\angle\text{EZB}=180^\circ\ .....(\text{i)}$
Also, $EF || BC$ and $EZ$ is the transversal.
$\Rightarrow\angle\text{BZE}=\angle\text{ZEF}$ (alternate angles)
$\Rightarrow\text{BZE}=\angle\text{DEF}\ .....(\text{ii)}$ From $(i)$ and $(ii)$,
we have $\angle\text{ABC}+\angle\text{DEF}=180^\circ$
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