# maths

in triangle abc angle b = 2angle c.d is a point ob bc such that ad bisects angle bac and ab = cd.prove that angle bac = 72

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1. make a sketch.
let angle C = x , then angle B = 2x
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:

sin(x+y)/sin2x = siny/sinx
sin(x + 90-3x/2)/(2sinxcosx) = siny/sinx
sin(90 - x/2)/(2cosx) = siny , after dividing by sinx
cross-multiply
sin(90-x/2) = 2sinxcosx
sin(90-x/2) = sin 2x
so 90 - x/2 = 2x
times 2
180 - x = 4x
180=5x
x=36
then y = 90-3(36)/2 = 36
and angle A = 2y = 72°

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2. 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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