Question:
In the given figure, AB || CD. If ∠CAB = 80° and ∠EFC = 25°, then ∠CEF = ?
(a) 65°
(b) 55°
(c) 45°
(d) 75°
Solution:
(c) 45°
$A B \| C D$ and $A F$ is the transversal.
$\therefore \angle D C F=\angle C A B=80^{\circ} \quad$ [Corresponding Angles]
Side EC of triangle EFC is produced to D.
$\therefore \angle C E F+\angle E F C=\angle D C F$
$\Rightarrow \angle C E F+25^{\circ}=80^{\circ}$
$\Rightarrow \angle C E F=55^{\circ}$
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