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

Given,
ABCD is a rhombus.DE is the altitude on AB then AE = EB.
In a ΔAED and ΔBED,
DE = DE (common line)
∠AED = ∠BED (right angle)
AE = EB (DE is an altitude)
∴ ΔAED
ΔBED (SAS property)
∴ AD = BD (by C.P.C.T)
But AD = AB (sides of rhombus are equal)
⇒ AD = AB = BD
∴ ABD is an equilateral triangle.
∴ ∠A = 60![]()
⇒ ∠A = ∠C = 60
(opposite angles of rhombus are equal)
But Sum of adjacent angles of a rhombus is supplementary.
∠ABC + ∠BCD = 180![]()
∠ABC + 60
= 180![]()
∠ABC = 180
-60
= 120![]()
∴ ∠ABC = ∠ADC = 120
(opposite angles of rhombus are equal)
∴ Angles of rhombus are ∠A = 60
, ∠C = 60
, ∠B = 120
, ∠D = 120![]()