Triangle ABC is an isosceles right triangle with a right angle at A. Segments BD and BE trisect angle ABC. What is the degree measure of angle BDE?
Since it is an isoc right triangle angle ABC (and ACB) are each 45 degrees . Trisecting the 45 degree angle results in three 15 degree angles . Angle ABD includes two of these so it is 30 degrees.
The resulting triangle (ABD) is 30 degrees + 90 degrees + 60 degrees. BDEA = 60 degrees.
Since ABC is isosceles, angles ACB and ABC = 45°
Since ABC is trisected, then angle DBE = 15°
Note that angle BEA = 90 - EBA = 90 - 15 = 75°
So....by the Exterior Angle Theorem , Angle BEA = angle DBE + angle BDE
75 = 15 + angle BDE subtract 15 from both sides
60° = angle BDE