A tangential quadrilateral is a quadrilateral whose four sides are all tangent to a circle inscribed within it. In such a quadrilateral, the sum of lengths of the two opposite sides of the quadrilateral is equal.

This is known as the Pitot theorem, named after Henri Pitot, a French engineer who proved it in the 18th century.

## Problem

A circle is inscribed in a quadrilateral ABCD. Show that |AB|+|CD| = |BC|+|DA|

## Strategy

We are working with a tangential quadrilateral, so all the sides are tangent to the circle. Let’s review the properties of tangents to a circle.

One of the theorems we have proved is the two tangent theorem – two tangents from the same point outside a circle have equal lengths to the points of tangency. This should be useful here, as it deals with the lengths of such tangents.

Looking at the quadrilateral, we have *four* such points outside the circle. Each one of the quadrilateral’s vertices is a point from which we drew two tangents to the circle.

Applying the two tangent theorem to each one of these points, we get 4 pairs of equal-size line segments: AP=AS=a; BP=BQ=b; CQ=CR=c and DR=DS=d.

And in each of the opposite pairs of sides of the quadrilateral, we have exactly one of these segments, so their sums are equal

## Proof

(1) AP=AS=a // Two tangent theorem

(2) BP=BQ=b // Two tangent theorem

(3) CQ=CR=c // Two tangent theorem

(4) DR=DS=d // Two tangent theorem

(5) |AB|+|CD| = |AP|+|PB|+|CR|+|RD|= a+b+c+d

(6) |BC|+|DA| = |BQ|+|QC|+|DS|+|SA|=b+c+d+a=a+b+c+d // Commutative property of addition

(7) |AB|+|CD| = |BC|+|DA| //(5),(6), transitive property of equality