# 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 $\angle XOM=\angle YOM.$ 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,

 $\angle DAB=\angle DCB$

and

 $\angle ADC=\angle ABC$

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 $\angle AKM=\angle CNM.$

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 $\angle AKM=\angle XOM$ and in $ONYM,$ we have $\angle CNM=\angle YOM.$ From these two, we get

 $\angle XOM=\angle YOM.$

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

Title proof of butterfly theorem ProofOfButterflyTheorem 2013-03-22 13:10:09 2013-03-22 13:10:09 drini (3) drini (3) 6 drini (3) Proof msc 51-00