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If ABCD is a convex quadrilateral in which $AB\times CD+AD \times BC=AC \times BD$ then ABCD is an inscribed quadrilateral

This is what I have so far

  • $\angle A +\angle B + \angle C + \angle D = 360$ because the total angles are 360 in a quadlateral
  • I would like to show that the opposite angles are supplementary so $\angle A$ and $\angle C$ are 180 or $\angle B$ and $\angle D$ are 180
  • Im am stuck on how to use the $AB\times CD+AD \times BC=AC \times BD$ and how to show one set of opposite angles are supplementary

Any help would be appreciated!!!

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Ptolemy's inequality is just the triangle inequality under circle inversion.

If $A,B,C,D$ (in this order) are the vertices of a convex quadrilateral, then $$ AC\cdot BD \leq AB\cdot CD + BC\cdot AD \tag{1}$$ holds. If we consider a unit circle centered at $A$ and the inversion with respect to such circle, bringing $B\mapsto B'$, $C\mapsto C'$, $D\mapsto D'$, then $(1)$ is equivalent to $$ B'D' \leq B'C'+C'D'\tag{2} $$ with equality holding only if $B',C',D'$ are collinear. It follows that equality in $(1)$ holds only for cyclic quadrilaterals.

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Here is how I prove both directions of Prolemy's theorem in one shot.

enter image description here

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

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  • $\begingroup$ @OliviaRyan What happened? Anything wrong with the proof? $\endgroup$ – Futurologist Apr 20 '17 at 1:34

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