Question

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

Answer 1

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 = 60^o

Since, opposite angles of rhombus are equal, we get,

⇒ ∠A = ∠C = 60^o

We also know that,

Sum of adjacent angles of a rhombus = supplementary.

So,

∠ABC + ∠BCD = 180^o

∠ABC + 60^o = 180^o

∠ABC = 180^o – 60^o = 120^o

Since, opposite angles of rhombus are equal, we get,

∠ABC = ∠ADC = 120^o

Hence, Angles of rhombus are:

∠A = 60^o, ∠C = 60^o, ∠B = 120^o, ∠D = 120^o