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In the figure, show that $AB \parallel EF$."
To do:
We have to show that $AB \parallel EF$.
Solution:
From the figure,
$\angle BAC = 57^o, \angle ACE = 22^o$
$\angle ECD = 35^o$ and $\angle CEF =145^o$
$\angle ECD + \angle CEF = 35^o + 145^o$
$ = 180^o$
$\angle ECD$ and $\angle CEF$ are co-interior angles.
Therefore,
$EF \parallel CD$
$\angle BAC = 57^o$
$\angle ACD = \angle ACE + \angle ECD$
$= 22^o+35^o$
$= 57^o$
This implies,
$AB \parallel CD$
$EF \parallel CD$ and $AB \parallel CD$
Therefore,
$AB \parallel EF$.
Hence proved.
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