Butterfly Theorem
The butterfly theorem is a well-known result from Euclidean geometry.
Contents
Statement
Let \(M\) be the midpoint of chord \(PQ\) of a circle.
Draw two other chords \(AB\) and \(CD\) through \(M\).
If \(AD\) and \(BC\) intersect chord \(PQ\) at \(X\) and \(Y\), respectively, \(M\) is the midpoint of \(XY\).
Looking at the diagram, you can probably tell how the butterfly theorem got its name!
Proof
There are various proofs for the butterfly theorem. We're going to prove it by using the properties of similar triangles.
We're going to start by dropping perpendiculars \(x_1\) and \(x_2\) from \(X\) to \(AB\) and \(CD\), respectively, and likewise \(y_2\) and \(y_1\) from \(Y\) to \(AB\) and \(CD\), respectively.
Here's what the diagram looks like after we do that:
Let \(MQ=PM=a\), \(XM=x\), and \(MY=y\).
Now we have a lot of similar triangles we can work with, and from these similar triangles we have a lot of ratios that are equal to one another:
\[\begin{array} &\dfrac{x}{y}=\dfrac{x_1}{y_2}, &\dfrac{x}{y}=\dfrac{x_2}{y_1}, &\dfrac{x_1}{y_1}=\dfrac{AX}{CY},& \dfrac{x_2}{y_2}=\dfrac{XD}{YB}.\end{array}\]
So, we have \(\dfrac{x^2}{y^2}=\dfrac{x_1}{y_2}\cdot \dfrac{x_2}{y_1}=\dfrac{x_1}{y_1} \cdot \dfrac{x_2}{y_2}=\dfrac{AX \cdot XD}{CY \cdot YB}\).
Now we're going to use the intersecting chords theorem. (Check the wiki out if you aren't familiar with it.)
From that theorem, we have, \(\dfrac{AX\cdot XD}{CY \cdot YB}=\dfrac{PX \cdot XQ}{PY \cdot YQ}\).
So, we can write \(\dfrac{x^2}{y^2}=\dfrac{PX \cdot XQ}{PY \cdot YQ}=\dfrac{(a-x)(a+x)}{(a+y)(a-y)}=\dfrac{a^2-x^2}{a^2-y^2}\).
In other words, \(\dfrac{x^2}{y^2}=\dfrac{a^2-x^2}{a^2-y^2}\). But that can only happen if \(x^2=y^2\).
This means \(x=y\), which implies \(M\) is the midpoint of \(XY\). \(_\square\)
On the circumference of circle \(\Gamma \), chord \(AB\) with length \(1100\) is drawn. Let \(C\) be the midpoint of \(AB\). Through \(C\), 2 other chords \( DE\) and \(FG\) are also drawn, such that the points around the circle are \(A, D, F, B, E, G\). The line segment \(AB\) intersects \(DG\) and \(FE\) (internally) at \(H\) and \(I\), respectively.
If \(AH=449\), what is \(CI?\)
Practice Problems
- Let the incircle \((\)with center \(I)\) of \(\triangle ABC\) touch the side \(BC\) at \(X\), and let \(A'\) be the midpoint of this side. Then prove that line \(A'I\) (extended) bisects \(AX\).
- Let \(PT\) and \(PB\) be two tangents to a circle, \(AB\) the diameter through \(B\), and \(TH\) the perpendicular from \(T\) to \(AB\). Then prove that \(AP\) bisects \(TH\).