Let $\angle ABC = x$, $\angle BCD = 2x$, and $\angle CDE = 3x$. If $\angle BAE$ is $90^{\circ}$, what is the angle of $\angle ABC$ in degrees?
arc BD = 4x
And angle DEA will intercept an arc of 90 + BD = 90 + 4x
So...its measure = (90 + 4x) / 2
And angle CDE = 3x
Let BC and DC intersect AE at F and G respectively
Angle FGC = 180 - [ angle DEA + angle CDE ] = 180 - [(90 + 4x) /2 + 3x]
Angle GFC = 90 - BCA = 90 - x
Angle BCD =2x
So ....in triangle GFC we that
Angle FGC + Angle BCD + Angle GFC =180
180 - [ (90 + 4x) / 2 + 3x ] + 2x + [ 90 - x ] = 180
180 - 45 - 2x - 3x + 2x + 90 - x = 180
-4x + 45 = 0
4x = 45
x = 45/4 = 11.25° = ABC