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?

Thanks!

AnonymousConfusedGuy
Feb 23, 2018

#1**+1 **

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.

ElectricPavlov
Feb 23, 2018

#2**+1 **

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

So

75 = 15 + angle BDE subtract 15 from both sides

60° = angle BDE

CPhill
Feb 23, 2018