Thread: No trigonometry use? triangle in a square.

ABCD is a square and ECD an isosceles triangle
with base angles 15 degrees, as shown in the figure. Prove that Angle AEB = 60 degrees (and therefore triangle AEB is equilateral), without using trigonometry.
Is this impossible? thanks for your help.

Originally Posted by Intsecxtanx

ABCD is a square and ECD an isosceles triangle
with base angles 15 degrees, as shown in the figure. Prove that Angle AEB = 60 degrees (and therefore triangle AEB is equilateral), without using trigonometry.
Is this impossible? thanks for your help.

I just did it. I got that AEB is equilateral.

Its easy to demonstrate that AEB can equilateral... but I have no idea how to prove this is the only available option.

Originally Posted by Intsecxtanx

ABCD is a square and ECD an isosceles triangle
with base angles 15 degrees, as shown in the figure. Prove that Angle AEB = 60 degrees (and therefore triangle AEB is equilateral), without using trigonometry.
Is this impossible? thanks for your help.

Angle AED = BEC. Angle DEC = 150. AED + BEC + DEC + AEB = 360.

Angle ADE = 75. Angle DAE = 90 - EAB. ADE + DAE + AED = 180.

I believe solving this system gives that the angles of the triangle are 60.