This is the proof that I have for Ptolemy's Theorem:
$\triangle ABC \sim \triangle ADM$ by AA so $\displaystyle\frac{AB}{AD} = \frac{AC}{AM} = \frac{BC}{DM}$
$\triangle ABD \sim \triangle ACM$ by AA so $\displaystyle\frac{AB}{AC} = \frac{AD}{AM} = \frac{BD}{CM}$
$CD= CM-DM = \displaystyle\frac{AC \cdot BD}{AB} – \frac{AD \cdot BC}{AB}$
From this we obtain $AB \cdot CD = AC \cdot BD – AD \cdot BC$
and $AC \cdot BD = AB \cdot CD + AD \cdot BC$.
Is there a way to use this proof of Ptolemy to formulate a proof for the converse?
Best Answer
You can prove both directions of Ptolemy's theorem:
On the same side of line $AC$ as point $D$, choose point $D^*$ so that $$\angle \, CAD^* = \angle \, BAD = \alpha+\tilde{\alpha} \,\,\, \text{ and } \,\,\, \angle \, ACD^* = \angle \, ABD = \beta$$ where $\angle \, BAC = \alpha$ and $\angle \, CAD = \tilde{\alpha}$, as drawn on the picture. Then triangles $ABD$ and $ACD^*$ are similar by construction which yields the identities $$\frac{AB}{AC} = \frac{AD}{AD^*} = \frac{BD}{CD^*}$$ The first and the last ratios yield $$CD^* = \frac{AC \cdot BD}{AB}$$ while the first identity $\frac{AB}{AC} = \frac{AD}{AD^*}$ can be reformulated as $$\frac{AB}{AD} = \frac{AC}{AD^*}$$ Combined with the fact that $$\angle \, BAC = \angle \, DAD^* = \alpha$$ because by construction $\angle \, CAD^* = \angle \, BAD = \alpha+\tilde{\alpha}$, it implies that triangles $ABC$ and $ADD^*$ are similar which, in it own turn, yields the identities $$\frac{AB}{AD} = \frac{AC}{AD^*} = \frac{BC}{DD^*}$$ Thus, one can write the expression $$DD^* = \frac{BC \cdot AD}{AB}$$ Therefore, based on the initial construction of point $D$, one can form the sum $$CD + DD^* = CD + \frac{BC \cdot AD}{AB} = \frac{AB \cdot CD + BC \cdot AD}{AB}$$
Proof of the converse of Ptolemy's Theorem. Assume that $$AB \cdot CD + BC \cdot AD = AC \cdot BD$$ Then $$CD + DD^* = \frac{AB \cdot CD + BC \cdot AD}{AB} = \frac{AC \cdot BD}{AB} = CD^*$$ which is possible if and only if the point $D$ lies on the line $CD^*$. Therefore, $$\angle \, ACD = \angle \, ACD^* = \angle ABD = \beta$$ which means that the quadrilateral $ABCD$ is cyclic.
Proof of Ptolemy's Theorem. Assume that $ABCD$ is cyclic quadrilateral. Then $$\angle \, ACD = \angle ABD = \beta$$ by cyclicity. But by construction $$\angle \, ACD^* = \angle \, ABD = \beta = \angle \, ACD $$ which is possible if and only if $D$ lies on the line $CD^*$. Therefore, $CD^* = CD + DD^*$ so by the already established expressions from the construction above $$\frac{AC \cdot BD}{AB} = CD^* = CD + DD^* = \frac{AB \cdot CD + BC \cdot AD}{AB}$$ which after canceling out the common denominator $AB$ turns into $$AB \cdot CD + BC \cdot AD = AC \cdot BD$$