Triangle.

Golden Ratio Triangle

|AB| =1 , |AD| =1 , |DB| =x , |CA| =1+x

if its a 72,72,36 triangle there is "Golden Ratio" so 1+x=1.618 .... x=0.618

We observe that $\angle ADB=72$ and $1=|AD|=|AB|$ .

Let $|BD|=x$

Therefore:

$\frac{1+x}{1}$ $=$ $\frac{1}{x}$

$\implies x^2+x-1=0$

Solving the quadratic yields:

$x_{1}=$ $\frac{-1+\sqrt5}{2}$

$x_{2}=$ $\frac{-1-\sqrt5}{2}$

The side can only take the positive value which is

$\phi-1\approx\boxed{0.618}$

Triangle CMD ~Triangle CAD CM/CA=MD/AB=1/CB 1/CA=MD/AB=1/1+ BD 1+BD=CA By using cos rule , cos 36=(CD^2 + AC^2 - AD^2)/2.CD.AC=(5^1/2 + 1)/2 BD=(5^1/2 - 1)/2 BD~0.618

Chasing down the angles, we see that triangle ACD is isoceles, and AD = CD =1. Applying the law of sines, AD/sin(72) = BD/sin(36), so BD = sin(36)/sin(72) = 0.618.

By the theorem of angle bisector $AB:AC=CD:BD$ and $AB=2AC×cosα$ substituting that we get $BD=0.618$

If the triangle is isosceles, each pink angle is 36º, and C is also 36º. Therefore, AD = 1. $\frac{1}{sen72º}$ = $\frac{?}{sen36}$ --> ? = 0.6180339887 ~ 0.618

Angle B is 72. so 1/2 angle A is 36, and angle C is 36. Triangle ACD is isosceles; since CD = 1, AD =1. Triangle ADB is also isosceles, and AB = 1.Let DB = d; then by the Law of Cosines in triangle ADB, we have d^2 = 1^2+ 1^2 - 2(1)(1) cos(36), d = .618. Ed Gray