# Chaitanya's Random Pages

## May 24, 2015

### A collection of proofs of Ptolemy’s Theorem

Filed under: mathematics — ckrao @ 11:39 am

Here we collect some proofs of the following nice geometric result.

If $ABCD$ is a quadrilateral, then $AB.CD + BC.DA \geq AC.BD$ with equality if $ABCD$ is cyclic.

In words, the sum of the product of the lengths of the opposite sides of a quadrilateral is at least the product of the lengths of its diagonals.

The equality for a cyclic quadrilateral is known as Ptolemy’s theorem while the more general inequality applying to any quadrilateral is called Ptolemy’s inequality. Many of the proofs below that establish Ptolemy’s theorem can be modified slightly to prove the inequality.

## Proofs of Ptolemy’s Theorem

Proof 1:

Choose point $P$ on line $CD$ extended beyond $D$ as shown so that $\angle DAP = \angle BAC$.

Then

a) $\triangle ABC \sim \triangle ADP$ ($\angle ABC = \angle ADP$ and $\angle BAC = \angle DAP$).

b) $\triangle BAD \sim \triangle CAP$ ($\angle BAD = \angle CAP$ and $\angle ADB = \angle APC$).

From a) $BC.DA = BA.DP$ so $AB.CD + BC.DA = AB(CD + DP) = AB.CP = AC.BD$ by b) and we are done.

Proof 2: (a slightly different choice of constructed point)

Let $K$ be the point on $AC$ so that $\angle CBK = \angle ABD$.

Then

a) $\triangle ABD \sim \triangle KBC$ ($\angle ABD = \angle KBC$ and $\angle BDA = \angle BCK$) so $KC = AD.BC/BD$.

b) $\triangle ABK \sim \triangle DBC$ ($\angle ABK = \angle DBC$ and $\angle KAB = \angle CDB$) so $AK = DC.AB/DB$.

Adding the two gives $AC = (AD.BC + DC.AB)/BD$ or $AC.BD = AD.BC + DC.AB$ as required.

Proof 3: (ref: http://mathafou.free.fr/themes_en/kptol.html)

Let the diagonals $AC, BD$ intersect at $P$ and construct $E$ on the circumcircle so that $CE$ is parallel to $BD$. A short angle chase shows that $BCED$ is an isosceles trapezium.

Then $\sin \angle ABE = \sin \angle ACE = \sin \angle BPC$ (angles on common arc $AE$ and using $BD \parallel CE$). Also triangles $BDE$ and $BDC$ have the same base and height and hence the same area.

Hence the area of ABCD is the sum of the areas of triangles ABE and ADE, or

\begin{aligned} & \frac{1}{2} \left( AB.BE\sin \angle ABE + AD.DE\sin \angle ADE \right)\\ &= \frac{1}{2} \left( AB.BE\sin \angle ABE + AD.DE\sin \angle ABE \right) \quad \text{as } \angle ABE \text{ and } \angle ADE \text{ are supplementary }\\ &= \frac{1}{2} \left( AB.BE+ AD.DE \right)\sin \angle ABE\\ &= \frac{1}{2} \left(AB.CD + AD.BC \right)\sin \angle ABE, \quad (1) \end{aligned}

where the last step follows from $BDEC$ being an isosceles trapezium.

But also this area is $\frac{1}{2} AC.BD\sin \angle BPC$ and recall from above that $\sin \angle BPC = \sin \angle ABE$. Therefore equating this with (1) we end up with $AB.CD + AD.BC = AC.BD$.

Proof 4 (ref: [1])

Here three of the triangles of the cyclic quadrilateral are scaled to fit together to form a parallelogram. Namely,

• $\triangle ABD$ is scaled by $AC$,
• $\triangle ABC$ is scaled by $AD$,
• $\triangle ACD$ is scaled by $AB$.

A simple angle chase based on the coloured angles shown reveals that a parallelogram is formed when the triangles are joined together. Ptolemy’s theorem then follows from equating two of its opposite side lengths.

Proof 5: (from here)

Using the cosine rule in triangles $ABC$ and $ADC$ respectively gives

$AC^2 = AB^2 + BC^2 - 2AB.BC\cos \angle ABC = AD^2 + CD^2 - 2AD.CD\cos \angle ADC.$

Since $\angle ABC + \angle ADC = 180^{\circ}$, this becomes

\begin{aligned} AB^2 + BC^2 - 2AB.BC\cos \angle ABC &= AD^2 + CD^2 + 2AD.CD\cos \angle ABC\\ \Rightarrow \cos \angle ABC &= \frac{AB^2 + BC^2 - AD^2 - CD^2}{2(AB.BC + AD.CD)}. \end{aligned}

Hence

\begin{aligned} AC^2 &= AB^2 + BC^2 - 2AB.BC\cos \angle ABC\\ &= AB^2 + BC^2 - 2AB.BC \frac{AB^2 + BC^2 - AD^2 - CD^2}{2(AB.BC + AD.CD)}\\ &= \frac{(AB^2+BC^2)(AB.BC + AD.CD) - AB.BC(AB^2 + BC^2 - AD^2 - CD^2)}{AB.BC + AD.CD}\\ &= \frac{(AB^2+BC^2)(AD.CD) + AB.BC(AD^2 + CD^2)}{AB.BC + AD.CD}\\ &= \frac{AB.AD(AB.CD+BC.AD) + BC.CD(BC.AD+AB.CD)}{AB.BC + AD.CD}\\ &= \frac{(AB.AD+BC.CD)(AB.CD+BC.AD)}{AB.BC + AD.CD}.\\ \end{aligned}

By a similar argument we may write

$\displaystyle BD^2 = \frac{(AB.BC + AD.CD)(BC.AD+AB.CD)}{BC.CD + AB.AD}$.

Multiplying these last two equations by each other and taking the square root gives

$\displaystyle AC.BD = AB.CD + BC.AD,$

which is Ptolemy’s theorem.

Proof 6 (via this):

Let the angles subtended by $AB, BC, CD$ respectively be $\alpha, \beta, \gamma$ and let the circumradius of the quadrilateral be $R$. By the sine rule $AB = 2R \sin \alpha, BC = 2R\sin \beta, CD = 2R \sin \gamma$, $AD = 2R\sin (\alpha + \beta + \gamma )$, $AC = 2R \sin (\alpha + \beta)$, $BD = 2R \sin (\beta + \gamma )$. Then the equality to be proved is equivalent to

$\displaystyle \sin (\alpha + \beta)\sin (\beta + \gamma) = \sin \alpha \sin \gamma + \sin \beta \sin (\alpha + \beta + \gamma).$

But we have

\begin{aligned} & \sin (\alpha + \beta)\sin (\beta + \gamma)\\ &= (\sin \alpha \cos \beta + \cos \alpha \sin \beta)(\sin \beta \cos \gamma + \cos \beta \sin \gamma)\\ &= \sin \alpha \sin \beta \cos \beta \cos \gamma + \sin \alpha \cos^2 \beta \sin \gamma + \cos \alpha \sin^2 \beta \cos \gamma + \cos \alpha \sin \beta \cos \beta \sin \gamma\\ &= \sin \alpha \sin \beta \cos \beta \cos \gamma + \sin \alpha (1- \sin^2 \beta ) \sin \gamma + \cos \alpha \sin^2 \beta \cos \gamma + \cos \alpha \sin \beta \cos \beta \sin \gamma\\ &= \sin \alpha \sin \gamma + \sin \beta (\sin \alpha \cos \beta \cos \gamma - \sin \alpha \sin \beta \sin \gamma + \cos \alpha \sin \beta \cos \gamma + \cos \alpha \cos \beta \sin \gamma )\\ &= \sin \alpha \sin \gamma + \sin \beta (\sin(\alpha + \beta ) \cos \gamma + \cos (\alpha + \beta ) \sin \gamma )\\ &= \sin \alpha \sin \gamma + \sin \beta \sin (\alpha + \beta + \gamma ), \end{aligned}

as required.

## Proofs of Ptolemy’s Inequality (all make use of the triangle inequality)

Proof 7:

Denote the points $A, B, C, D$ by vectors or complex numbers $a, b, c, d$. Then we have the equality

$\displaystyle (a-b).(c-d) + (a-d).(b-c) = (a-c).(b-d).$

Applying the modulus (length) to both sides and then the triangle inequality leads to

$\displaystyle |(a-b).(c-d)| + |(a-d).(b-c)| \geq |(a-c).(b-d)|,$

which is Ptolemy’s inequality.

Proof 8: (ref: [2])

Place the origin at the point $D$ so that $A, B, C$ are represented by the vectors $a, b, c$ respectively. Let $a' = a/|a|^2, b' = b/|b|^2, c' = c/|c|^2$. Then

\begin{aligned} |a'-b'|^2 &= \left| \frac{a}{|a|^2} - \frac{b}{|b|^2}\right|^2\\ &= \frac{|a|^2}{|a|^4} + \frac{|b|^2}{|b|^4} - 2\frac{\langle a, b \rangle}{|a|^2|b|^2}\\ &= \frac{|b|^2 + |a|^2 - 2\langle a, b \rangle}{|a|^2|b|^2}\\ &= \frac{|a-b|^2}{|a|^2|b|^2}. \end{aligned}

Similarly, $|b'-c'| = \frac{|b-c|}{|b||c|}$ and $|c'-a'| = \frac{|c-a|}{|c||a|}$. By the triangle inequality, $\displaystyle |a'-b'| \leq |b'-c'| + |c'-a'|$, or in other words,

\begin{aligned} \frac{|a-b|}{|a||b|} &\leq \frac{|b-c|}{|b||c|} + \frac{|c-a|}{|c||a|}\\ \Rightarrow |a-b||c| \leq |b-c||a| + |c-a||b|, \end{aligned}

which is another way of writing Ptolemy’s inequality.

Proof 9: (inversion)

Recall that under inversion under a point $P$ a circle passing through $P$ maps to a line. If $X, Y$ are points on the circle mapping to $X, Y'$ respectively under inversion in a circle of radius $R$ centred at $P$, then

$\displaystyle X'Y' = R^2.XY/(PX.PY).$

To see this, since $XP.X'P = YP.Y'P = R^2$ by the definition of the inverse, $XP/YP = Y'P/X'P$. This together with the fact that angle $P$ is common implies that triangles $XPY$ and $Y'PX'$ are similar. This gives us the relationship $X'Y' = YX.PY'/PX = XY.R^2/(PX.PY)$ as desired.

For our quadrilateral $ABCD$ apply an inversion centred at $D$ with radius $R$. Then $A, B, C$ map to points $A', B', C'$ which are collinear if $ABCD$ is cyclic. Using the result above $A'B' = AB.R^2/(DA.DB)$ and similarly $A'C' = AC.R^2/(DA.DC)$ and $B'C' = BC.R^2/(DB.DC)$. By the triangle inequality $A'C' \leq AB' + BC'$ and so this becomes

\begin{aligned} \frac{AC.R^2}{DA.DC} &\leq \frac{AB.R^2}{DA.DB} + \frac{BC.R^2}{DB.DC}\\ \Rightarrow AC.DB &\leq AB.DC + BC.DA, \end{aligned}

which is Ptolemy’s inequality.

Proof 10: (ref: [3])

Construct $E$ on $AC$ so that $EC = BC$, then draw $F$ on $CD$ with $EF \parallel AD$. Finally rotate $\triangle FEC$ about $C$ to map to $\triangle F'BC$.

Then $F'B = FE$ and by similarity of triangles $ACD$ and $ECF$, $FE/DA = CE/CA = CB/CA$ so that

(a) $F'B = FE = AD.BC/CA$

Secondly, $\triangle DF'C \sim ABC$ as $F'C/BC = FC/EC = DC/AC$ and $\angle F'CD = \angle BCA$. This gives

(b) $DF' = AB.CD/AC$

Adding (a) and (b) and applying the triangle inequality,

$\displaystyle BD \leq BF' + F'D = (AD.BC + AB.CD)/AC,$

from which

$\displaystyle AC.BD \leq AD.BC + AB.CD$

which is Ptolemy’s inequality.

#### References

[1] A. Bogomolny, Ptolemy Theorem – Proof Without Words from Interactive Mathematics Miscellany and Puzzles
http://www.cut-the-knot.org/proofs/PtolemyTheoremPWW.shtml, Accessed 30 May 2015

[2] W.H. Greub, Linear Algebra, Springer Science & Business Media, 2012 (p 190).

[3] C. Alsina, R. B. Nelsen, When Less is More: Visualizing Basic Inequalities, MAA, 2009.