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

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°

♥️I HOPE U ALL ENJOY THIS ANSWER ♥️ THANK U

## Answers ( )

Answer:THE ANSWERS SEE IN GOOGLE

Step-by-step explanation:SEE IN GOOGLE

AND

I AM LOVER OF FREE FIRE