Question

ABCD is a rhombus in which the altitude from D to side AB bisects AB. Find the angles of the rhombus. [4 MARKS]

Answer 1

Concept: 2 Marks
Application: 2 Marks

Let DL be the altitude from D on side AB. It is given that L bisects AB i.e.

AL = BL. Join points B and D to form diagonal BD.
In $\Delta$ ADL and $\Delta$BDL,

DL = DL [Common side]
$\angle ALD=\angle BLD={90}^{\circ }$ [Given]
AL = BL [Given]

So, by SAS congruence criterion,
$\Delta ADL\cong \Delta BDL$
$\Rightarrow AD=BD$

Since all sides of a rhombus are equal in length,
$AD=CD=BC=AB=BD$
$\Rightarrow △ABD$ and $△BCD$ are equilateral.

$\Rightarrow \angle A=\angle C={60}^{\circ }$

$\angle ABC=\angle ABD+\angle DBC={60}^{\circ }+{60}^{\circ }={120}^{\circ }$
$\angle ADC=\angle ABC={120}^{\circ }$ [Opposite angles of a parallelogram are equal.]