Consider ΔABC with BC as base. AB = BC = CA
Draw a line PQ through A parallel to BC.
Then, angle PAB = angle ABC (alternate interior angles) - 1
Similarly, angle QAC = angle ACB - 2
angle PAB + angle BAC + angle CAQ = 180 (sum of angles on a line)
Then, from 1 and 2,
angle ABC + angle BAC + angle ACB = 180
angle ABC = BAC = ACB (equilateral triangle)
3 * any one of these angles equal = 180
each angle = 180/3 = 60
Hope it is enough and you understood.