In geometry, Pitot's theorem describes the relation between the opposite sides of a tangential quadrilateral.
It was named after the French engineer, Henri Pitot who proved it in \(1725\).
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.
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!
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.
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 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 \(CD\).
1) Circle \( \Gamma \) lies outside of quadrilateral \( ABCD \). If \(AB, BC, CD, DA \) extended are tangential to \( \Gamma\), show that \( AB + CD = BC + DA \).
2) Prove the converse to the above problem.
3) 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 \(BD \).