Question
In Fig. $\text{BA}||\text{ED}$ and $\text{BC}||\text{EF}$ . Show that $\angle\text{ABC}=\angle\text{DEF}$
[Hint: Produce $DE$ to intersect $BC$ at $P$ (say)].
[Hint: Produce $DE$ to intersect $BC$ at $P$ (say)].
✓
Answer
Produce $DE$ to intersect $BC$ at P(say). $\text{EF}||\text{BC}$ and $DP$ is the transversal,
$\therefore\angle\text{DEF}=\angle\text{DPC}.....(1)$ $[\text{Corres}.\angle\text{S}]$
Now, $\text{AB}||\text{DP}$ and $BC$ is the transversal, $\therefore\angle\text{DPC}=\angle\text{ABC}..(2)$ $[\text{Corres}.\angle\text{S}]$ From $(1)$ and $(2),$ we get
$\angle\text{ABC}=\angle\text{DEF}$ Hence, Proved.
$\therefore\angle\text{DEF}=\angle\text{DPC}.....(1)$ $[\text{Corres}.\angle\text{S}]$
Now, $\text{AB}||\text{DP}$ and $BC$ is the transversal, $\therefore\angle\text{DPC}=\angle\text{ABC}..(2)$ $[\text{Corres}.\angle\text{S}]$ From $(1)$ and $(2),$ we get
$\angle\text{ABC}=\angle\text{DEF}$ Hence, Proved.
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