Ptolemy's Theorem/Proof 1

Theorem

Let $ABCD$ be a cyclic quadrilateral.

Then:

$AB \times CD + AD \times BC = AC \times BD$

Proof

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:

$AC \times BD = AB \times CD + BC \times DA$

$\blacksquare$