Segment $BD$ and $AE$ intersect at $C$, as shown, $AB=BC=CD=CE$, and $\angle A = \frac 32 \angle B$. What is the degree measure of $\angle D$?
+128406 Since AB = BC....then angles BAC and BCA are equal
Call angle ABC = x
Then angles BAC and BCA = (3/2) x
So
(3/2)x + (3/2) x + x =180
3x + x =180
4x = 180
x = 180 / 4 = 45°
So angle BCA = (3/2) 45 = 67.5 = angle ECD
And because CD = CE....then angles CED and CDE are equal
So
Angle CDE = (180 - 67.5) / 2 = 56.25° = "angle D"
