Question

In a rhombus ABCD, the altitude from D to side AB bisects AB. Find the angles of the rhombus.

Answer 1

Given that ABCD is a Rhombus and 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 traingle.

∴ ∠A = 60°

⇒ ∠A = ∠C = 60° (opposite angles of rhombus are equal)

But Sum of adjacent angles of a rhombus is supplimentary.

∠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° and ∠C = 60° , ∠B = ∠D = 120°.

Answer 2

Given:

A rhombus ABCD

altitude D bisects AB

To Find:

angles of the rhombus

Solution:

In rhombus ABCD,

AE = EB [DE is an altitude]

∠AED = ∠BED = 90°

Now, we join BD,

In ΔAED and ΔBED

AE = BE [given]

∠DEA = ∠BED [ 90° angle]

DE = DE [common]

ΔAED ≅ ΔBED [by SAS congruency criteria]

AD = BD [C.P.C.T] ..(i)

Since, ABCD is a rhombus

AD = AB ..(ii)

Now, using (i) and (ii),

AD = BD = AB

ΔABD is an equilateral triangle

So,

∠A = ∠B = ∠D = 60°

AB ║DC and BC is transversal

So, ∠B + ∠C = 180°

60° + ∠C = 180°

∠C = 120°

As we know that opposite angles of a rhombus are equal

So, ∠A = ∠C = 120°

∠B = ∠D = 60°

Therefore, the angles of the rhombus are ∠A = 60° and ∠C = 60°, ∠B = ∠D = 120°.