In this diagram, the triangle ABC is isosceles, since the two sides, AB and AC, are equal as are the two angles at B and C.
[Also, angles ABC and ACB are twice angle CAB.]

If we bisect the base angle at B by a line from B to point D on AC then we have the angles as shown and also angle BDC is also 72°. BCD now has two angles equal and is therefore an isosceles triangle; and also we have BC=BD.

Since ABD also has two equal angles of 36°, it too is isosceles and so BD=AD. So in the diagram the three sides BC, BD and DA are all the same length.

We also note that the little triangle BCD and the whole triangle ABC are similar since they are both 36°-72°-72° triangles.

Let's call the smallest segment here, CD, length 1 and find the lengths of the others in relation to it. We will therefore let the ratio of the smaller to longer sides in triangle BCD be r so that if CD is 1 then BC is r.

In the larger triangle ABC, the base is now r and as it is the same shape as BCD, then its sides are in the same ratio so Ab is r times BC, e.d. AB is r².
Also, we have shown BC=BD=AD so AD is r (and CD is 1).
From the diagram we can see that AC=AD+DB.
But AC=1+r and in isosceles triangle ABC, AB (which is r²) is the same length as AC (which is 1+r), so
r² = 1 + r
and this is the equation which defined the golden ratio.

So the triangle with angles 36°, 72°, 72° has sides that are proportional to Phi, Phi and 1 (which is the same as 1,1,phi).

Pentagrams and the 36°-72°-72° triangle

If we look at the way a pentagram is constructed, we can see there are lots of lines divided in the golden ratio: Since the points can be joined to make a pentagon, the golden ratio appears in the pentagon also and the relationship between its sides and the diagonals (joining two non-adjacent points).

The reason is that Phi has the value 2 cos (π/5) where the angle is described in radians, or, in degrees, Phi=2 cos (36°).
[See below for more angles whose sines and cosines involve Phi!]

Since the ratio of a pair of consecutive Fibonacci numbers is roughly equal to the golden section, we can get an approximate pentagon and pentagram using the Fibonacci numbers as lengths of lines:
There is another flatter triangle inside the pentagon here. Has this any golden sections in it? Yes! We see where further down this page, but first, a quick and easy way to make a pentagram without measuring angles or using compasses:

Making a Paper Knot to show the Golden Section in a Pentagon

Here's an easy method to show the golden section by making a Knotty Pentagram; it doesn't need a ruler and it doesn't involve any maths either!
Take a length of paper from a roll - for instance the type that supermarkets use to print out your bill - or cut off a strip of paper a couple of centimetres wide from the long side of a piece of paper. If you tie a knot in the strip and put a strong light behind it, you will see a pentagram with all lines divided in golden ratios.

This is my favourite method since it involves a Knot(t)!

Here are 5 pictures to help (well it is a pentagram so I had to make 5 pictures!) - although it really is easy once you practice tying the knot!

1. As you would tie a knot in a piece of string ...
2. ... gently make an over-and-under knot, rolling the paper round as in the diagram.
3. (This is the slightly tricky bit!)
   Gently pull the paper so that it tightens and you can crease the folds as shown to make it lie perfectly flat.
4. Now if you hold it up to a bright light, you'll notice you almost have the pentagram shape - one more fold reveals it ...
5. Fold the end you pushed through the knot back (creasing it along the edge of the pentagon) so that the two ends of the paper almost meet. The knot will then hang like a medal at the end of a ribbon.
   Looking through the knot held up to the light will show a perfect pentagram, as in the diagram above.