Ptolemy's Theorem via a bit of Trig and similar triangles

Here’s a straightforward trigonometric proof of Ptolemy’s Theorem using scale factors and the Law of Cosines. First we start with a cyclic quadrilateral and assign it segment lengths:

1. $ \triangle{CDE} $ has edges of length x, y and z.
2. $ \triangle{BEA} \sim \triangle{CDE} $ so we give a scale factor k and then edges of length kx, ky, kz.
3. $ \triangle{BEC} \sim \triangle{AED} $ and from the given values the scale factor is x:y so we add the last side w and scale it appropriately to wy and wx.

Next we compute $ AD \cdot BC + CD \cdot AB = w^2xy + kz^2 $ And then we apply the Law of Cosines twice on the central point E

• $ w^2 = k^2 + 1 - 2k\cos{(\pi - \theta)} = k^2 + 1 + 2k\cos{\theta} $
• $ z^2 = x^2 + y^2 - 2xy\cos{\theta} $

(Note for the second one I scaled everything down.) Finally we substitute these expression into the original one which leads to a happy cancellation and we get:

\(AD \cdot BC + CD \cdot AB = (k^2 + 1 + 2k\cos{\theta}) \cdot xy + k \cdot (x^2 + y^2 - 2xy\cos{\theta})\)\(AD \cdot BC + CD \cdot AB = k^2xy + xy + kx^2 + ky^2\)\(AD \cdot BC + CD \cdot AB = (ky + x)\cdot(kx + y)\)\(AD \cdot BC + CD \cdot AB = AC \cdot BD\)