**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.**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)

*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 $V$ arereflectedthrough $M$, 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.)

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

**Exercise 2.**(

*School Science and Mathematics*, 1919)

- 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 $$\frac{MQ^2}{MP^2} = \frac{QB \times QC}{PA \times PD}.$$
- 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)

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

**Exercise 4.**(W.G. Horner,

*The Gentleman's Diary*, 1815)

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

**Exercise 5.**(Leon Bankoff,

*School Science and Mathematics*, 1955)

- Prove that $AML$ is an isosceles triangle.
- Prove that $MLCQ$ is a cyclic quadrilateral.
- 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.

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