+0  
 
+1
199
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avatar+1442 

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
avatar+12735 
+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
edited by ElectricPavlov  Feb 23, 2018
 #2
avatar+88899 
+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

 

 

cool cool cool

CPhill  Feb 23, 2018
 #3
avatar+1442 
+1

Thanks so much!

AnonymousConfusedGuy  Feb 24, 2018

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