According to the question,
We have,
ABCD is a rhombus.
DE is the altitude on AB then AE = EB.
In ΔAED and ΔBED,
We have,
DE = DE (common line)
∠AED = ∠BED (right angle)
AE = EB (DE is an altitude)
∴ ΔAED ≅ ΔBED by 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 = 60o
Since, opposite angles of rhombus are equal, we get,
⇒ ∠A = ∠C = 60o
We also know that,
Sum of adjacent angles of a rhombus = supplementary.
So,
∠ABC + ∠BCD = 180o
∠ABC + 60o = 180o
∠ABC = 180o – 60o = 120o
Since, opposite angles of rhombus are equal, we get,
∠ABC = ∠ADC = 120o
Hence, Angles of rhombus are:
∠A = 60o, ∠C = 60o, ∠B = 120o, ∠D = 120o