Question

In the given figure, AB || DC, ∠BAD = 90°, ∠CBD = 28° and ∠BCE = 65°. Then ∠ABD = ?
(a) 32°
(b) 37°
(c) 43°
(d) 53°

Answer 1

Side DC of triangle BCD is produced to E.
$$\begin{gathered} \therefore \angle BCE=\angle CBD+\angle BDC \\ \Rightarrow 65°=28°+\angle BDC \\ \Rightarrow \angle BDC=37° \end{gathered}$$
Now,
$AB\parallel DC\text{and}BD$ is the transversal.
$$\begin{gathered} \angle ABD=\angle BDC=37°\left[\text{Alternate Interior Angles}\right] \\ \therefore \angle ABD=37° \end{gathered}$$