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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