Question

In the given figure, BAD || EF, ∠AEF = 55° and ∠ACB = 25°. Find ∠ABC.

Answer 1

$BAD\parallel EF\text{and}AE$ is the transversal.
$$\begin{gathered} \therefore \angle AEF+\angle EAD=180°\left[\text{Consecutive Interior Angles}\right] \\ \Rightarrow 55°+\angle EAD=180° \\ \Rightarrow \angle EAD=125° \end{gathered}$$
Also,
$\angle BAC=\angle EAD=125°\left[\text{Vertically-Opposite Angles}\right]$
From $∆ABC$, we have:
$$\begin{gathered} \angle BAC+\angle ABC+\angle ACB=180°\left[\text{Sum of the angles of a triangle}\right] \\ \Rightarrow 125°+\angle ABC+25°=180° \\ \Rightarrow \angle ABC=30° \end{gathered}$$