make a sketch.
let angle C = x , then angle B = 2x
also, let angle BAD = y = angle CAD , (Angle A was bisected by AD)
then in triangle ABC , 2x + 2y + x = 180
y = (180-3x)/2 = 90 - 3x/2
by exterior angle theorem,
angle BDA = x+y and angle CDA = 2x+y
let AB = m = DC , (given)
by the sine law:
m/sin(x+y) = AD/sin 2x and m/siny = AD/sinx
m = ADsin(x+y)/sin 2x and m = ADsiny/sinx
so ADsin(x+y)/sin2x = ADsiny/sinx
divide by AD
sin(x+y)/sin2x = siny/sinx
sin(x + 90-3x/2)/(2sinxcosx) = siny/sinx
sin(90 - x/2)/(2cosx) = siny , after dividing by sinx
sin(90-x/2) = 2sinxcosx
sin(90-x/2) = sin 2x
so 90 - x/2 = 2x
180 - x = 4x
then y = 90-3(36)/2 = 36
and angle A = 2y = 72°
There is a wonderful field of math lying ahead from this problem.
notice that cos 36° = .809016.. which is 1/2 of the golden ratio of (1+√5)/2 or 1.61803....
Look up the "pentagon"
Draw the central triangles, your triangle is one of these.
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