1. ## 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

2. 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 $\displaystyle arc~AB$ Let $\displaystyle P$ be a point on the arc such that $\displaystyle AP = BC$ note that $\displaystyle APBC$ is an isosceles trapzeium so we have $\displaystyle AB = CP$ and $\displaystyle S_{ABC} = S_{ACP}$ . Then consider the area of the quadrilateral $\displaystyle ABCD$ we find that it is equal to :

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Here is another method ,

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

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

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

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

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

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

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

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

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

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

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

$\displaystyle ac+bd= mn$

3. LOL, uhmm, wheres all the numbers gone after this website upgrade?? can you rewrite this again?? sorry! thanks!