Question

In the adjoining figure, ABCD is a square ans triangle EDC is an equilateral triangle. Prove that
(i) AE = BE
(i) angle DAE = 15°​

Answer 1

i- ED=EC

AD=BC

<EDC+<ADC = <ECD+<BCD or <EDA = <ECB

Hence, Triangle ADE and Triangle BCE are equal.

Since, AE = BE as Triangle ADE and Triangle BCE

are equal.

ii- As DE = DC = AD, hence DE = AD

If DE is equal to AD then angle DAE is equal to

angle DEA. Let them as (y).

Hence y + y + 90°+60° = 180°

2y= (180-90-60)°

2y = 30°

y = 30/2 = 15°

Angle DAE is equal to 15°