Triangle ABC is an isosceles triangle with AC=BC and angle C =120°. Points D and E are chosen on segment AB so that AD=DE=EB. Find angle CDE
+128406 C = 120
A = 30 D M E B = 30
x x x
Let M be the midpoint of AB
Then angle ACM = 60 angle AMC = 90 angle CAM = 30
Then ACM is A 30 - 60-90 right triangle
AB = 3x
AM = (1/2) AB = (3/2)x
AM - AD = DM
(3/2)x - x = DM
x/2 = DM
CM = AM / sqrt 3 = (3/2)x / sqrt (3) = [sqrt (3) / 2] x
tan (angle CDM) = CM /DM
tan ( angle CCM) = [(sqrt (3) / 2) x ] / ( x / 2)
tan (angle CDM) = sqrt 3
arctan (sqrt (3) ) = angle CDM = angle CDE = 60°
