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

Answers (1)

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

Here $\angle DLA=\angle DLB=90^{\circ}$ {DL is perpendicular bisector of AB}
$AL=BL=\frac{X}{2}$
DL = DL {common side of $\triangle ADL$ and $\triangle BDL$ }
$\triangle ALD\cong \triangle BLD$ {by SAS congruence}
$AD = BD$
In $\triangle ADB$ , $AD = AB = DB = x$
$\triangle ADB$ is an equilateral triangle.
$\therefore \angle ADB=\angle ABD=\angle A=60^{\circ}$
Similarly, $\triangle DCB$ is an equilateral triangle
$\angle BDC=\angle DBC=\angle C=60^{\circ}$
Also, $\angle A = \angle C$
$\angle D = \angle B$ …..(1)
$\angle A+ \angle B+\angle C+\angle D=360^{\circ}$