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

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°

♥️I HOPE U ALL ENJOY THIS ANSWER ♥️ THANK U​

in progress 0
Liliana 1 month 2021-08-17T14:22:53+00:00 1 Answer 0 views 0

Answers ( )

  1. Charlotte
    0
    2021-08-17T14:24:37+00:00

    Answer:

    THE ANSWERS SEE IN GOOGLE

    Step-by-step explanation:

    SEE IN GOOGLE

    AND

    I AM LOVER OF FREE FIRE

Leave an answer

Browse
Browse

18:9+8+9*3-7:3-1*13 = ? ( )