**Pascal's hexagon theorem**and the

**butterfly theorem**.

First, let us talk about the

*butterfly theorem*. The butterfly theorem is stated as follows.Butterfly Theorem.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. Then the point $M$ is the midpoint of $PQ$.

We can see that the picture of the butterfly theorem looks like a

*butterfly*with*two wings*joining at the point $M$. That's probably the reason why it bears this name.
There have been many proofs discovered for the butterfly theorem. These proofs employ quite a wide range of techniques. Some proofs use Menelaus' theorem, some proofs make use of the concept of power of a point with respect to a circle, or radical axis, yet, some other proofs use trigonometry, analytic geometry, etc...

Today, we will present a simple proof of the butterfly theorem by using

**Pascal's hexagon theorem**. Pascal's theorem is stated as follows.Pascal's Theorem.Given a hexagon $123456$ inscribed in a circle. Then the intersection points of the three pairs of the opposite sides of the hexagon $$\{12, 45\}, ~\{23, 56\}, ~\{34, 61\}$$ are collinear.

Pascal discovered this amazing theorem when he was only 16 years old. An interesting aspect of this theorem is that it has many configurations. The six vertices of the hexagon can be arranged on the circle in any particular order. So with each arrangement of the vertices, we have an instance of the Pascal's theorem.

Pascal's theorem is also known as the magic hexagon theorem. Instead of inscribing a hexagon on a circle, we can inscribe a hexagon on any conic -

*ellipse*,*parabola*, or*hyperbola*- the theorem is still true! How magical is that?!
Proving the butterfly theorem

Let us now prove the butterfly theorem by using Pascal's hexagon theorem.

Draw the diameters $AU$ and $CV$. Suppose $DU$ and $BV$ meet at $N$. Apply Pascal's theorem for the hexagon $ABVCDU$, we deduce that the three points $M$, $O$, $N$ lie on a straight line.

Since $M$ is the midpoint of the chord $XY$, the line $NOM$ is perpendicular to $XY$. Since $AU$ and $CV$ are diameters, we have $$\angle ADU = \angle CBV = 90^{o}.$$ It follows that $MNDP$ and $MNBQ$ are two inscribed quadrilaterals. Thus, $$\angle MNP = \angle MDP = \angle MBQ = \angle MNQ .$$ It follows that the two right triangles $MNP$ and $MNQ$ are congruent, from here we obtain the required equality $MP=MQ$.

So today, we have found a new application of Pascal's theorem. The picture used in the proof portrays a beautiful union between

*Pascal's hexagon theorem*and the*butterfly theorem*. That's why I have chosen a name for it. I call it**Pascal's butterfly**.
Hope to see you again in the next post.

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