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 ZBAD = ZCAE = 15°,
find the angles of the isosceles triangle.
15°
15°
B
D-
E
С​

Answer 1

Answer:

Given ABC is an isosceles triangle with AB=AC D and E are the point on BC such that BE=CD

Given AB=AC

∴∠ABD=∠ACE (opposite angle of sides of a triangle ) (1)

Given BE=CD

Orbc =ce (2)

In ΔABD and ΔACE

∠ABD=∠ACE ( From 1)

BC=CE (from 2)

AB=AC ( GIven)

∴ΔABD≅ΔACE

So AD=AE [henceproved]