Question

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





In ∆ AED AND ∆BED,
DE= DE (COMMON)
Angle AED = Angle BED (Each having 90°)
AE = BE [GIVEN]
=> ∆ AED IS CONGRUENT TO ∆ BED [SAS CONGRUENCE RULE]
=> AD = BD [BY CPCT]

But, AD = AB [Sides of a rhombus are equal]
=>AD = AB= BD => ∆ BAD is equilateral triangle.
=> Angle A = 60°
Also, angle A = angle C = 60° [Opp. angles of rhombus are equal]
Now, Angle ABC + angle BCD = 180°
=> Angle ABC = 120°
Now, angle ABC = angle ACD = 120° [Opp. angles of rhombus are equal]

So, Angle A= 60°
Angle B = 120°
Angle C = 60°
Angle D = 120°


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