Lesson · Tangent lengths and pitot
The incircle and its relatives
A circle touching the three side lines of a triangle produces six tangent segments in three equal pairs, and those lengths obey bookkeeping so tight that most tangency problems become arithmetic. This chapter sets the bookkeeping up once - for the incircle, for tangential quadrilaterals, and for the excircles - and then records the two configurations built on top of it that national rounds keep returning to: the touch chord and the arc midpoint.
Tangent lengths and Pitot's theorem
Theorem
Let the incircle of a triangle \(ABC\) with sides \(a = BC\), \(b = CA\), \(c = AB\) and semiperimeter \(s = \tfrac{1}{2}(a+b+c)\) touch \(BC\), \(CA\), \(AB\) at \(D\), \(E\), \(F\). Then
Indeed the tangent segments from each vertex are equal, say \(x\) from \(A\), \(y\) from \(B\), \(z\) from \(C\); then \(y + z = a\), \(z + x = b\), \(x + y = c\), adding all three gives \(x + y + z = s\), and subtracting one equation at a time finishes it.
Principle
Do not chase touch points; chase the three letters. Once every touch point carries its tangent length, any statement about lengths along the sides is a linear identity in \(s-a\), \(s-b\), \(s-c\), and it can be checked - or discovered - by adding equations.
The same bookkeeping in a quadrilateral has a name.
Theorem
(Pitot) If a convex quadrilateral \(ABCD\) has an inscribed circle, then
with tangent lengths \(x, y, z, w\) at the four vertices, both sides equal \(x + y + z + w\). Conversely, every convex quadrilateral with \(AB + CD = BC + DA\) admits an inscribed circle. The converse is true but harder, and we use it without proof; it is the direction that produces a circle out of a length identity.
Excircles
Definition
The excircle of \(ABC\) opposite \(A\) is the circle tangent to the side \(BC\) and to the extensions of \(AB\) and \(AC\) beyond \(B\) and \(C\). Its centre \(I_A\) lies on the internal bisector from \(A\) and on the external bisectors from \(B\) and \(C\); there is one excircle opposite each vertex.
Theorem
The tangent length from \(A\) to the excircle opposite \(A\) equals \(s\).
Let the excircle touch the line \(AB\) at \(P\), the line \(AC\) at \(Q\) and the side \(BC\) at \(D'\). Equal tangents from \(B\) give \(BP = BD'\), equal tangents from \(C\) give \(CQ = CD'\), so
and \(AP = AQ\) forces both to equal \(s\).
So, measured from a vertex, the incircle answers \(s - a\) and the opposite excircle answers \(s\): the two most quotable lengths of the subject. Constant perimeter is the excircle's signature - if a triangle cut off from an angle has perimeter \(2s\), its excircle opposite the vertex touches the two fixed arms at distance \(s\) from it, no matter what the cutting line did.
The touch chord and the arc midpoint
Lemma
(Touch chord) Let the incircle touch \(AB\) at \(F\) and \(AC\) at \(E\), and let \(\alpha = \angle BAC\). The chord \(EF\) makes the angle \(90^\circ - \tfrac{\alpha}{2}\) with each of the two sides and is perpendicular to the bisector \(AI\).
The triangle \(AEF\) is isosceles because \(AE = AF = s - a\), so its base angles are \(\tfrac{1}{2}\left(180^\circ - \alpha\right) = 90^\circ - \tfrac{\alpha}{2}\); and its axis of symmetry is the bisector from \(A\), which therefore cuts \(EF\) at a right angle.
The point of the lemma is that the touch chord has a known direction. A line whose angle with every side is computable meets midlines, bisectors and parallels at computable angles - which is how a hypothesis "the incircle touches here" turns into an angle chase.
Lemma
(Arc midpoint) Let \(M\) be the midpoint of the arc \(BC\) of the circumcircle of \(ABC\) not containing \(A\), and let \(I\) be the incentre. Then \(M\) lies on the bisector from \(A\) and
Equal arcs \(BM\) and \(MC\) are cut off by equal chords and seen from \(A\) under equal inscribed angles, so \(MB = MC\) and \(AM\) bisects \(\angle BAC\). For the third equality chase two angles: \(\angle MBI = \angle MBC + \angle CBI = \tfrac{\alpha}{2} + \tfrac{\beta}{2}\), since \(\angle MBC = \angle MAC\) over the arc \(MC\); and \(\angle MIB\), the exterior angle of the triangle \(ABI\) at \(I\), equals \(\tfrac{\alpha}{2} + \tfrac{\beta}{2}\) as well. The triangle \(MBI\) is isosceles, so \(MI = MB\).
Read forwards, the lemma hands you a circle centred at \(M\) through \(B\), \(C\) and \(I\). Read backwards, it locates the incentre: of the two points where that circle meets the line \(AM\), the incentre is the one on the \(A\)-side of \(M\) - and the other one is the excentre \(I_A\), which satisfies \(MI_A = MB\) too (true, not proved here). The whole family gathers on one circle.
Warning
The midpoint in the lemma is the midpoint of the arc \(BC\) not containing \(A\). The midpoint of the other arc lies on the external bisector from \(A\), and it is the excentres \(I_B\) and \(I_C\) that it holds at chord distance, not the incentre.
Warm-up
Example
The incircle of a triangle \(ABC\) touches the side \(BC\) at \(D\), and the excircle opposite \(A\) touches \(BC\) at \(D'\). Prove that \(BD = CD'\), so that \(D\) and \(D'\) are mirror images of each other in the midpoint of \(BC\).
One of the two distances is already on the shelf
The incircle bookkeeping gives \(BD = s - b\) at once. Everything therefore depends on computing \(CD'\), a tangent length of the excircle from \(C\).
Measure the excircle from \(A\) first
The tangent length from \(A\) to the excircle is \(s\), reaching the touch point on the line \(AC\) beyond \(C\). The piece of it beyond \(C\) has length \(s - b\), and the equal tangent segments from \(C\) carry that length over to the side \(BC\).
Compare and reflect
So \(CD' = s - b = BD\). Then also \(BD' = a - CD' = s - c = CD\), and the midpoint of \(BC\) is the midpoint of \(DD'\): reflecting the side in its midpoint swaps the two touch points.