In the figure, arms $BA$ and $BC$ of $\angle ABC$ are respectively parallel to arms $ED$ and $EF$ of $\angle DEF$. Prove that $\angle ABC = \angle DEF$. "
Given:
In the figure, arms $BA$ and $BC$ of $\angle ABC$ are respectively parallel to arms $ED$ and $EF$ of $\angle DEF$.
To do:
We have to prove that $\angle ABC = \angle DEF$.
Solution:
Produce $BC$ to meet $DE$ at $G$
$AB \parallel DE$
This implies,
$\angle ABC = \angle DGH$.........…(i) (Corresponding angles)
$BH \parallel EF$
$\angle DGH = \angle DEF$................(ii) (Corresponding angles)
From (i) and (ii), we get,
$\angle ABC = \angle DEF$.
Hence proved.
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