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.

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