Ptolemy's Theorem/Proof 2

Theorem

Let $ABCD$ be a cyclic quadrilateral.

Then:

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

Proof

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Let $\Box ABCD$ be a cyclic quadrilateral, with diagonals $AC$ and $BD$.

By Opposite Angles of Cyclic Quadrilateral sum to Two Right Angles:

$\angle ABC$ is supplementary to $\angle ADC$

As well:

$\angle BAD$ is supplementary to $\angle BCD$

Construct two triangles $\triangle A'B'C'$ and $\triangle C'D'E'$ congruent to $\triangle ABC$ and $\triangle CDE$ respectively, with $B'C'D'$ collinear.

Although this article appears correct, it's inelegant. There has to be a better way of doing it. In particular: This is messy. The new triangles have different points, so we can't refer to them as $\triangle ABC$ and $\triangle CDE$. We also need to be more precise about exactly what we are doing, because currently it's woolly. You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by redesigning it. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Improve}} from the code. If you would welcome a second opinion as to whether your work is correct, add a call to {{Proofread}} the page.

$E$ is the same point that was $A$ in Figure $1$.

Hence:

$ED = AD$

$\angle ABC$ and $\angle CDE$ are supplementary.

By Equal Corresponding Angles or Supplementary Interior Angles implies Parallel Lines:

$AB \parallel ED$

By construction:

$AC = EC$

Now, scale the sides of $\triangle CDE$ by the length of $AB$.

Also scale the sides of $\triangle ABC$ by the length of $DE$.

Let the scaled figure be $A'B'F'G'$.

$A'B' = AB \cdot DE = F'G'$

By Quadrilateral is Parallelogram iff One Pair of Opposite Sides is Equal and Parallel:

$\Box A'B'F'G'$ is a parallelogram

By construction:

$\angle A'C'B' + \angle A'C'F'$ has the same measure of angle as the original $\angle BCD$.

Therefore the supplementary angle $\angle A'C'G'$ has the same measure as the supplementary angle $\angle BAD$.

By construction:

$A'C' = AC \times DE$

$B'C' = BC \times DE$

Also by construction:

$C'G' = CE \times AC = AD \cdot AC$

$C'F' = CD \times AC$

Therefore $A'C'$ and $C'G'$ are in proportion with scale factor $AC$.

By Triangles with One Equal Angle and Two Sides Proportional are Similar:

$\triangle A'C'H' \sim \triangle ABD$

Thus, the length of $A'G'$ of the new construct is equal to $BD$ times the scale factor $AC$ Euclid:Proposition/VI/4:

$A'G' = AC \times BD$

And:

$B'F' = B'C' + C'F'$

Substituting:

$B'F' = BC \times DE + CD \times AC$

But by Opposite Sides and Angles of Parallelogram are Equal:

$A'G' = B'C'F' = B'F'$

Therefore

$AC \times BD = BC \times DE + CD \times AC$

$\blacksquare$