In the given figure, AB || CD. If ∠CAB = 80° and ∠EFC = 25°, then ∠CEF = ?


In the given figure, AB || CD. If CAB = 80° and ∠EFC = 25°, then ∠CEF = ?
(a) 65°
(b) 55°
(c) 45°
(d) 75°


(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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