Ptolemy's Theorem

Theorem
Let $ABCD$ be a cyclic quadrilateral.

Then:
 * $AB \times CD + AD \times BC = AC \times BD$

Proof

 * PtolemysTheorem.png

Let $ABCD$ be a cyclic quadrilateral.

By Angles in Same Segment of Circle are Equal:
 * $\angle BAC = \angle BDC$

and:
 * $\angle ADB = \angle ACB$

By Construction of Equal Angle, construct $E$ on $AC$ such that:
 * $\angle ABE = \angle CBD$

Since:
 * $\angle ABE + \angle CBE = \angle ABC = \angle CBD + \angle ABD$

it follows that:
 * $\angle CBE = \angle ABD$

By Equiangular Triangles are Similar:
 * $\triangle ABE$ is similar to $\triangle DBC$

and:
 * $\triangle ABD$ is similar to $\triangle EBC$

Thus:
 * $\dfrac {AE} {AB} = \dfrac {CD} {BD}$

and:
 * $\dfrac {CE} {BC} = \dfrac {DA} {BD}$

Equivalently:
 * $AE \times BD = AB \times CD$

and:
 * $CE \times BD = BC \times DA$

Adding:
 * $AE \times BD + CE \times BD = AB \times CD + BC \times DA$

Factorizing:
 * $\paren {AE + CE} \times BD = AB \times CD + BC \times DA$

But:
 * $AE + CE = AC$

so:
 * $AC \times BD = AB \times CD + BC \times DA$