proof of butterfly theorem


Given that M is the midpointMathworldPlanetmathPlanetmathPlanetmath 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 perpendicularMathworldPlanetmathPlanetmathPlanetmath 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 XOM=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,

DAB=DCB

and

ADC=ABC

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

ADAM=BCCM

or

AKKM=CNNM

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 AKM=CNM.

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

In OKXM, we have AKM=XOM and in ONYM, we have CNM=YOM. From these two, we get

XOM=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