Question

ABCD is a rhombus such that the perpendicular bisector of AB passes through D. Find the angles of the rhombus.

Hint: Join BD. Then ∆ABD is equilateral.

Answer 1

Let ABCD be a rhombus where DE is perpendicular bisector of AB.

Construction: Join BD.

Now, in triangle AED and triangle BED:

AE = EB

ED = ED  ...[Common side]

∠AED = ∠DEB = 90° 

Now, using SAS rule,

ΔAED ≅ ΔBED

AD = DB = AB   ...[ABCD is a rhombus. So, AD = AB]

Hence, triangle ADB is an equilateral triangle.

So, ∠DAB = ∠DBA = ∠ADB = 60°

∴ ∠DCB = 60°   ...[Opposite angles of a rhombus are equal]

Now, ∠DAB + ∠ABC = 180°  ...[Adjacent angles of a rhombus are supplementary]

⇒ 60° + ∠ABD + ∠DBC = 180°

⇒ 60° + 60° + ∠DBC = 180°

⇒ ∠DBC = 60°

⇒ ∠ABC = ∠ABD + ∠DBC = 60° + 60° = 120°

∴ ∠ADC = 120°  ...[Opposite angles of a rhombus are equal]

Hence, the angles of the rhombus are 60°, 120°, 60°, 120°.