Question

4. Within an equilateral triangle ABC, an isosceles
triangle ADE is drawn, where D and E are points
on BC such that AD = AE. If angle BAD = angle CAE = 15,
find the angles of the isosceles triangle.​

Answer 1

Angle B A D equals to 15 degree

angle EAC is equals to 15

degree

angle BAC is equal to 60 degree (angle of a triangle is 60 degree )

then angle BAC is equals to angle BAD+ angle DAE+ angleEAC

60 is equals to 15 + angle DAE + 15

60 - 30 is equals to angle DAE

DAE=30

angle ADE is equals to angle A E D angles opposite to equal sides are equal

angle A D E + angle AED + angleDAE is equal to 180

angle A D E + angle ADE+30=180

2ADE=150

ADE=75

AED=75

DAE=30

Answer 2

Step-by-step explanation:

Angle B A D equals to 15 degree

angle EAC is equals to 15

degree

angle BAC is equal to 60 degree (angle of a triangle is 60 degree)

then angle BAC is equals to angle BAD+ angle DAE+ angleEAC

60 is equals to 15 + angle DAE + 15

60-30 is equals to angle DAE

DAE=30

angle ADE is equals to angle A E D

angles opposite to equal sides are equal

angle A D E + angle AED + angleDAE is

equal to 180 angle A D E + angle ADE+30=180

2ADE÷150

ADE=75

AED=75

DAE=30

brainless please