Question
In the given figure, $BA || ED$ and $BC || EF$. Show that $\angle\text{ABC}=\angle\text{DEF}.$
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Answer
Construction: Produce $DE$ to meet $BC$ at $Z$.
Now, $AB || DZ$ and $BC$ is the transversal.
$\Rightarrow\angle\text{ABC}=\angle\text{DZC}$ $($corresponding angles$) ….(i)$
Also, $EF || BC$ and $DZ$ is the transversal.
$\Rightarrow\angle\text{DZC}=\angle\text{DEF}$ $($corresponding angles$) ….(ii)$
From $(i)$ and $(ii)$, we have
$\angle\text{ABC}=\angle\text{DEF}$
Now, $AB || DZ$ and $BC$ is the transversal.
$\Rightarrow\angle\text{ABC}=\angle\text{DZC}$ $($corresponding angles$) ….(i)$
Also, $EF || BC$ and $DZ$ is the transversal.
$\Rightarrow\angle\text{DZC}=\angle\text{DEF}$ $($corresponding angles$) ….(ii)$
From $(i)$ and $(ii)$, we have
$\angle\text{ABC}=\angle\text{DEF}$
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