Question

Question 179
ABCD is a rhombus such that the perpendicular bisector of AB passes through D. Find the angles of the rhombus.
[Hint Join BD. Then, $\Delta ABD$ is equilateral]

Answer 1

Let ABCD be a rhombus in which DE is perpendicular bisector of AB.

Join BD. Then, in $\Delta AED$ and $\Delta BED$, we have
AE = EB
ED = ED [common side]
$\angle AED=\angle DEB={90}^{\circ }$
Then, by SAS rule, $\Delta AED\cong BED$
$\therefore$ AD = DB = AB [$\because$ ABCD is a rhombus. So, AD = AB]
Thus, $\Delta ADB$ is an equilateral triangle.
$\therefore \angle DAB=\angle DBA=\angle ADB={60}^{\circ }$
$\Rightarrow \angle DCB={60}^{\circ }$ [opposite angles of a rhombus are equal]
Now, $\angle DAB+\angle ABC={180}^{\circ }$ [adjacent angles of a rhombus are supplementary]
$\Rightarrow {60}^{\circ }+\angle ABC={180}^{\circ }\Rightarrow \angle ABC={180}^{\circ }-{60}^{\circ }={120}^{\circ }$
$\therefore \angle ADC={120}^{\circ }$ [opposite angles of a rhombus are equal]
Hence, the angles of the rhombus are ${60}^{\circ },{120}^{\circ },{60}^{\circ },{120}^{\circ }$.