# Pitot's Theorem

In geometry, **Pitot's theorem** describes the relationship between the opposite sides of a tangential quadrilateral. The theorem is a consequence of the fact that two tangent line segments from a point outside the circle to the circle have equal lengths. There are four equal pairs of tangent segments, and both sums of opposite sides can each be decomposed into sums of these four tangent segments. The converse is also true: a circle can be inscribed into every convex quadrilateral in which the lengths of opposite sides sum to the same value.

It was named after the French engineer Henri Pitot who proved it in 1725.

## Statement

In a tangential quadrilateral (a quadrilateral in which a circle can be inscribed), the two sums of lengths of the opposite sides are equal.

Let \(ABCD\) be a tangential quadrilateral. Then Pitot's theorem tells us that \(AB+CD=AD+BC\).

The converse of this is also true, but requires quite a bit more work. It was only proved in 1846 by Jakob Steiner, more than 100 years after Pitot published his proof!

## Proof

Pitot's theorem is an immediate consequence of a well-known fact that the lengths of the two tangent line segments from a point outside a circle to the circle are equal.

We're going to prove this here.

What we have to prove is \(BC=BD\).

Note that both triangles \(ABC\) and \(ABD\) are right angled.

We also have \(AC=AD\) (since they are the radii of the circle).

\(AB\) is a side that is common to both \(\triangle ABC\) and \(\triangle ABD\).

This means that the triangles are congruent and we can say that \(BC=BD.\) \(_\square\)

Now Pitot's theorem should be obvious. The picture below sums it up quite nicely.

## Proof of Converse

We want to show that in a convex quadrilateral \(ABCD\), if \(AB+CD = BC + DA \), then we can inscribe a semicircle that is tangential to all \(4\) sides.

Note that we can always draw a circle tangent to \(AB\), \(BC\) and \(CD\). The center of that circle is the point the angle bisectors of \(\angle ABC\) and \(\angle BCD\) intersect at. This point always exists and so does the circle.

All we have to do is to show that this circle is tangent to \(DA\) as well.

We're going to do a proof by contradiction.

Assume that the circle is not tangent to \(AD\). Now draw a tangent to the circle from \(A\) and let \(E\) be the point that it intersects with \(CD\).

Here's a picture where \(E\) lies in the interior of \(CD\).

Since the circle is inscribed in the quadrilateral \(ABCE\), from Pitot's theorem we have \(AB + CE =AE+BC\). Remember that we had \(AB+CD = BC + DA \) or \(AB + CE + ED = AD + BC\). But that implies \(ED + AE = AD\), which is impossible since \(\triangle AED\) is non-degenerate.

So our previous assumption was wrong and the circle is tangential to all \(4\) sides of \(ABCD\). \(_\square\)

The proof would be almost the same even if \(E\) were not in the interior of \(CD\).

## Additional Problems

- Circle \( \Gamma \) lies outside of quadrilateral \( ABCD \). If \(AB, BC, CD, DA \) extended are tangential to \( \Gamma\), show that \( AB + CD = BC + DA \).
- Prove the converse to the above problem.
- Show that if \( AB + CD = BC + DA \), then the incircles of \( ABC \) and \( BCD \) are tangential.

In fact, all 4 incircles of \( ABC, BCD, CDA, DAB \) are concurrent at the intersection of \(AC\) and \(BD \).