In the given figure,$AB \parallel DE$ and $BC \parallel EF$. Prove that $\angle ABC = \angle DEF$.
(N/A) Given: $AB \parallel DE$ and $BC \parallel EF$.
To prove: $\angle ABC = \angle DEF$.
Construction: Extend $DE$ to intersect $BC$ at point $G$.
Proof:
$1$. Since $AB \parallel DE$ and $BC$ is a transversal,$\angle ABC = \angle DGC$ (Corresponding angles).
$2$. Since $BC \parallel EF$ and $DE$ is a transversal,$\angle DGC = \angle DEF$ (Corresponding angles).
$3$. From equations $(1)$ and $(2)$,we get $\angle ABC = \angle DEF$.
Hence proved.
To prove: $\angle ABC = \angle DEF$.
Construction: Extend $DE$ to intersect $BC$ at point $G$.
Proof:
$1$. Since $AB \parallel DE$ and $BC$ is a transversal,$\angle ABC = \angle DGC$ (Corresponding angles).
$2$. Since $BC \parallel EF$ and $DE$ is a transversal,$\angle DGC = \angle DEF$ (Corresponding angles).
$3$. From equations $(1)$ and $(2)$,we get $\angle ABC = \angle DEF$.
Hence proved.
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