# Ptolemy's Theorem

• Jun 9th 2010, 06:24 PM
cedricc
Ptolemy's Theorem
In a cycic quadrilateral ABCD, let the sides AB, BC, CD, DA be of lengths a, b, c, d, respectively.

If the diagonal AC and BD have lengths m and n, respectively, prove that ac + bd = mn
• Jun 9th 2010, 08:13 PM
simplependulum
So you need a proof of Ptolemy's Theorem , right ?

Here's what i proved not long ago :

There are fours arcs , choose the biggest one , say $arc~AB$ Let $P$ be a point on the arc such that $AP = BC$ note that $APBC$ is an isosceles trapzeium so we have $AB = CP$ and $S_{ABC} = S_{ACP}$ . Then consider the area of the quadrilateral $ABCD$ we find that it is equal to :

$\frac{1}{2} (AM \cdot BM + BM \cdot CM + CM \cdot DM + DM \cdot AM ) ~\sin(\theta)$ where $M$ is the intersection of the diagonals and $\theta$ is the included angle (acute one ) .

$= \frac{1}{2} \sin(\theta) ( AM+CM)(BM+DM)$

$= \frac{mn}{2} \sin(\theta)$

Since $S_{ABC} = S_{ACP}$ , we have

$S_{ABC} + S_{ACD} = S_{ACP} + S_{ACD}$

$S_{ABCD} = S_{ACP} + S_{ACD}$

Also $S_{ACP} + S_{ACD} = S_{APCD} = S_{ADP} + S_{CDP}$

$S_{ADP} = \frac{1}{2} AD \cdot AP \sin(\angle DAP)$

$= \frac{1}{2} AD \cdot BC \sin(\angle DAP) ~,~ (AP = BC) ~,
$
$= \frac{1}{2} bd \sin(\angle DAP)$

$S_{CDP} = \frac{1}{2} CD \cdot CP \sin(\angle DCP)$

$= \frac{1}{2} CD \cdot AB \sin(\angle DAP) ~ ,~AB=CP ~ ,~\angle DCP = 180^o - \angle DAP ~,$ $= \frac{1}{2} ac \sin(\angle DAP)$

But $\angle DAP = \angle CAD + \angle PAC = \angle CAD + \angle ACB = \angle CAD + \angle ADB = 180^o - \theta$

Therefore, $S_{ADP} + S_{CDP} = \frac{1}{2} (ac+bd) \sin(\theta)$

By equating $S_{ABCD}$ and $S_{ADP} + S_{CDP}$ we have $mn = ac + bd$

Here is another method ,

Let $\alpha = \angle ACB ~,~ \beta = \angle ACD ~,~ x = \angle BAC ~,~ y = \angle CAD$

$ac+bd = (2R\sin(\alpha))(2R \sin(y)) + (2R\sin(\beta))(2R\sin(x))$

$= 4R^2 [ \sin(\alpha) \sin(y) + \sin(\beta)\sin(x)]$

$= 2R^2 [ \cos(\alpha-y)- \cos(\alpha+y) + \cos(\beta - x)- \cos(\beta + x ) ]$

$= 2R^2 [ (\cos(\alpha-y) + \cos(\beta - x) ) - ( \cos(\alpha+y) + \cos(\beta + x ) )]$

$= 4R^2 [ \cos\left( \frac{\alpha-y+\beta - x}{2} \right)\cos\left( \frac{\alpha-y-\beta + x}{2} \right)$ $- \cos\left( \frac{\alpha+y+\beta + x}{2} \right) \cos\left( \frac{\alpha+y-\beta - x}{2} \right) ]$

Since $\alpha+\beta +x+y = 180^o$

$= 4R^2 [\cos(90^o - (\alpha+\beta)) \cos( 90^o - (\alpha + x)) - \cos(90^o) \cos\left( \frac{\alpha+y-\beta - x}{2} \right) ]$

$= 4R^2 [ sin(\alpha+\beta) \sin(\alpha + x ) -0 ]$

$= (2R sin(\alpha+\beta) ) (2R \sin(180^0 -(\alpha + x )))$

$= (2R \sin(\angle BCD))(2R \sin(\angle ABC) ) = mn$

$ac+bd= mn$
• Jun 18th 2010, 02:14 AM
cedricc
LOL, uhmm, wheres all the numbers gone after this website upgrade?? can you rewrite this again?? sorry! thanks!