Question
In Fig.$\text{BA}||\text{ED}$ and $\text{BC}||\text{EF}$ Show that $\angle\text{ABC}+\angle\text{DEF}=180^\circ$
✓
Answer
Produce $ED$ to meet $BC$ at $P($say$)$
Now, $\text{EF}||\text{BC}$ and $EP$ is the transversal.
$\angle\text{DEF}+\angle\text{EPC}=180^\circ....(1)$
Again, $\text{EP}||\text{AB}$ and $BC$ is the transversal.
$\therefore\angle\text{EPC}=\angle\text{ABC}....(2)$ $[\text{corresponding}\angle\text{S}]$
From $(1)$ and $(2),$ we get $\angle\text{DEF}+\angle\text{ABC}=180^\circ$
$\Rightarrow\angle\text{ABC}+\angle\text{DEF}=180^\circ$ Hence, proved.
Now, $\text{EF}||\text{BC}$ and $EP$ is the transversal.
$\angle\text{DEF}+\angle\text{EPC}=180^\circ....(1)$
Again, $\text{EP}||\text{AB}$ and $BC$ is the transversal.
$\therefore\angle\text{EPC}=\angle\text{ABC}....(2)$ $[\text{corresponding}\angle\text{S}]$
From $(1)$ and $(2),$ we get $\angle\text{DEF}+\angle\text{ABC}=180^\circ$
$\Rightarrow\angle\text{ABC}+\angle\text{DEF}=180^\circ$ Hence, proved.
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