The butterfly problem

In the previous post, we show a simple proof of the butterfly problem by using Pascal's hexagon theorem. Today, we will list a few more proofs.  These proofs will be presented in the form of exercises so that interested readers can brush up on their problem-solving skills .

First, let us state the butterfly problem.

Butterfly Problem. In a circle $(O)$, let $M$ be the midpoint of a chord $XY$. Draw two chords $AB$ and $CD$ passing through $M$. Suppose the two lines $AD$ and $BC$ meet $XY$ at $P$ and $Q$, respectively. Prove that $M$ is the midpoint of $PQ$.

We can see that the picture of the butterfly problem looks like a butterfly with two wings joining at the point M. That's probably the reason why it bears this name.

The butterfly problem has many proofs. We select a few elementary proofs from the following two papers and presented them here in the form of exercises. These exercises are suitable for high school students.

• Leon Bankoff, The metamorphosis of the butterfly problem, Mathematics Magazine, vol. 60, no. 4, Oct 1987, p. 195-210.

• Greg Markowsky, Pascal's hexagon theorem implies the butterfly theorem, Mathematics Magazine, vol. 84, no. 1, Feb 2011, p. 56-62.

Exercise 1. (Greg Markowsky, Mathematics Magazine, 2011)

Suppose $IJ$ is a diameter of the circle $(O)$ and $M$ is a point on $IJ$. Two points $U$ and $V$ on the circle are called reflected through $M$ if they are on the same side of the line $IJ$ and $\angle IMU = \angle JMV$. Prove the following lemma.

Lemma. If $U$ and $V$ are reflected through $M$, and $Z$ is the intersection point of $IV$ and $JU$, then $ZM$ is perpendicular to $IJ$. 

(Hint. Suppose $UM$, $VM$ meet the circle at $V'$, $U'$; and $IV'$ intersects $JU'$ at $Z'$. Prove that $U'$, $V'$, $Z'$ are reflection of $U$, $V$, $Z$ across the line $IJ$. Use Pascal's hexagon theorem to prove that $Z$, $M$, $Z'$ lie on a straight line.)

Through $M$, draw the diameter $IJ$. Let $E$ and $F$ be the reflection points of $A$ and $D$ across the line $IJ$. Let $K$ be the intersection point of $IF$ and $JC$, $L$ the intersection point of $IB$ and $JE$, $Q'$ the intersection point of $BC$ and $EF$.

1. Prove that $C$ and $F$ are reflected through $M$; $B$ and $E$ are reflected through $M$. 
2. Use the lemma to prove that $M$, $K$, $L$ lie on $XY$. 
3. Use Pascal's hexagon theorem to prove that $K$, $L$, $Q'$ lie on $XY$.
4. Prove that $Q = Q'$, and deduce $MP = MQ$.

Exercise 2. (School Science and Mathematics, 1919)

Let $N$ be the intersection point of $AD$ and $BC$.

1. Use Menelaus' theorem for the triangle $NPQ$ with the following triples $\{C, M, D\}$, $\{A, M, B\}$, and use the concept of power of a point with respect to a circle, prove that $$\frac{MQ^2}{MP^2} = \frac{QB \times QC}{PA \times PD}.$$
2. Prove that $$\frac{MQ^2}{MP^2} = \frac{MY^2 - MQ^2}{MX^2 - MP^2},$$ and deduce $MP = MQ$.

Exercise 3. (Richard Taylor, The Gentleman's Diary, 1815)

Suppose the circumcircle of the triangle $APM$ intersects with the circle $(O)$ at $T$; the lines $TP$, $TM$ intersect the circle $(O)$ at $E$, $F$, respectively.

1. Prove that the three lines $EB$, $XY$, $DF$ are parallel.
2. Prove that the two triangles $MTD$ and $MCF$ are congruent.
3. Prove that the two triangles $MTP$ and $MCQ$ are congruent and deduce $MP = MQ$.

Exercise 4. (W.G. Horner, The Gentleman's Diary, 1815)

Let $I$, $J$ be the midpoints of $AD$ and $BC$.

1. Prove that $MBC$ and $MDA$ are two similar triangles.
2. Prove that $\angle MJQ = \angle MIP$.
3. Prove that the two triangles $MOP$ and $MOQ$ are congruent and deduce that $MP = MQ$.

Exercise  5. (Leon Bankoff, School Science and Mathematics, 1955)

Through $A$, draw a line parallel to $XY$, which meets the circle at $L$.

1. Prove that $AML$ is an isosceles triangle.
2. Prove that $MLCQ$ is a cyclic quadrilateral.
3. Prove that the two triangles $MAP$ and $MLQ$ are congruent and deduce that $MP = MQ$.

Let us stop here for now. There are many more proofs of the butterfly problem. Curious readers are referred to Leon Bankoff's paper listed above. Hope to see you again next time.