proof of butterfly theorem

Given that $M$ is the midpoint of a chord $PQ$ of a circle and $AB$ and $CD$ are two other chords passing through $M$, we will prove that $M$ is the midpoint of $XY,$ where $X$ and $Y$ are the points where $AD$ and $BC$ cut $PQ$ respectively.

Let $O$ be the center of the circle. Since $OM$ is perpendicular to $XY$ (the line from the center of the circle to the midpoint of a chord is perpendicular to the chord), to show that $XM=MY,$ we have to prove that $\mathrm{\angle }XOM=\mathrm{\angle }YOM.$ Drop perpendiculars $OK$ and $ON$ from $O$ onto $AD$ and $BC$, respectively. Obviously, $K$ is the midpoint of $AD$ and $N$ is the midpoint of $BC$. Further,

$$\mathrm{\angle }DAB=\mathrm{\angle }DCB$$

and

$$\mathrm{\angle }ADC=\mathrm{\angle }ABC$$

as angles subtending equal arcs. Hence triangles $ADM$ and $CBM$ are similar and hence

$$\frac{AD}{AM}=\frac{BC}{CM}$$

or

$$\frac{AK}{KM}=\frac{CN}{NM}$$

In other words, in triangles $AKM$ and $CNM,$ two pairs of sides are proportional. Also the angles between the corresponding sides are equal. We infer that the triangles $AKM$ and $CNM$ are similar. Hence $\mathrm{\angle }AKM=\mathrm{\angle }CNM.$

Now we find that quadrilaterals $OKXM$ and $ONYM$ both have a pair of opposite straight angles. This implies that they are both cyclic quadrilaterals.

In $OKXM,$ we have $\mathrm{\angle }AKM=\mathrm{\angle }XOM$ and in $ONYM,$ we have $\mathrm{\angle }CNM=\mathrm{\angle }YOM.$ From these two, we get

$$\mathrm{\angle }XOM=\mathrm{\angle }YOM.$$

Therefore $M$ is the midpoint of $XY.$

Title	proof of butterfly theorem
Canonical name	ProofOfButterflyTheorem
Date of creation	2013-03-22 13:10:09
Last modified on	2013-03-22 13:10:09
Owner	drini (3)
Last modified by	drini (3)
Numerical id	6
Author	drini (3)
Entry type	Proof
Classification	msc 51-00