Question

Question 35
In the given figure, ABCD and BDCE are parallelograms with common base DC. If $BC\perp BD$, then $\angle BEC$ is equal to
a) ${60}^{\circ }$
b) ${30}^{\circ }$
c) ${150}^{\circ }$
d) ${120}^{\circ }$

Answer 1

$\angle BAD={30}^{\circ }$ [given]
$\therefore$ $\angle BCD={30}^{\circ }$ [$\because$ opposite angles of a parallelogram are equal]
In $\Delta CBD$, by angle sum property of a traingle, we have,
$\angle DBC+\angle BCD+\angle CDB={180}^{\circ }\Rightarrow {90}^{\circ }+{30}^{\circ }+\angle CDB={180}^{\circ }\Rightarrow \angle CDB={180}^{\circ }-{120}^{\circ }={60}^{\circ }$
$\therefore \angle BEC=\angle CDB={60}^{\circ }$ [$\because$ opposite angles of a parallelogram are equal]