In a given figure abc is an equilateral triangle and bcde is a square prove that angle Abe equals 15 degree?

Assuming the configuration is as follows. With the triangle inside the square D B A E C Angle CBD = 90 & Angle ABC = 60, therefore angle ABD = 30 (90-60) We also know that angle DBE = 45 (half square) So DBE-ABD=ABE which is 45 - 30 = 15 Its a bit confusing without proper diagrams but it works.