Question

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

Answer 1

Let sides of a rhombus be AB = BC = CD = DA = x

Now, join DB.

In $\Delta ALD\text{and}\Delta BLD$, $\Delta DLA=\Delta DLB={90}^{\circ }$

[Since, DL is a perpendicular bisector of AB]

$AL=BL=\frac{x}{2}$

and DL=DL [Common side]


$\therefore \Delta ALD=\Delta BLD$ [By SAS congruence rule]

AD = BD [ By CPCT]

Now, in $\Delta ADB$ is an equilateral triangle.

$\therefore \angle C=\angle BDC=\angle DBC={60}^{\circ }$

Also, $\angle A=\angle C$

$\therefore \angle D=\angle B={180}^{\circ }-{60}^{\circ }={120}^{\circ }$ [Since , sum of interior angles is ${180}^{\circ }$]