# An Extension of van Schooten's Theorem

van Schooten's Theorem asserts that, for a point P on the circumcircle of an equilateral ΔABC, the longest of the segments PA, PB, PC is the sum of the shorter two.

Bui Quang Tuan has shown how van Schooten theorem can be extended to triangles that are not equilateral.

### Theorem 1

Given ΔABC and a point P on its circumcircle, a, b, c denote the sides of ABC and da, db, dc the distances from P to sides a = BC, b = CA, c = AB. We are going to show that one of the ratios a/da, b/db, c/dc is the sum of the other two.

The proof depends on the following

### Lemma

In ΔABC, h = b·c/(2R), where h is the altitude from A and R is the circumradius. (For the proof see Metric Relations in a Triangle.)

To continue with the theorem, for ΔPBC, lemma gives

da = PB·PC/(2R)

and similarly for triangles PCA and PAB. Next we obtain

 (1) a/da = a/(PB·PC)·(2R) = a·PA·(2R)/(PA·PB·PC), b/db = b/(PC·PA)·(2R) = b·PB·(2R)/(PA·PB·PC), c/dc = c/(PA·PB)·(2R) = c·PC·(2R)/(PA·PB·PC).

ΔABC divides its circumcircle into three arcs, say (BC), (CA), (AB). If, for instance, P is on arc (BC) then by Ptolemy's theorem,

a·PA = b·PB + c·PC.

Therefore

 (2) a/da = b/db + c/dc,

as needed.

It follows from (1) that say, when a = b = c, PA = PB + PC is equivalent to a/da = b/db + c/dc. However, while the latter holds for all triangles, the former holds only for the equilateral ones.

Observe that, as it follows from the proof, the ratio in the left-hand side of (2) (i.e., the largest ratio possible) corresponds to the arc that contains P.

Now, as Bui Quang Tuan has shown, (2) admits a further extension to polygons with more than three vertices.

### Theorem 2

Given point P on the circumcircle of a cyclic convex polygon X1X2X3 ... Xn-1Xn (n > 2), we use following notations (I and j can be interchanged):

• ai j = length of side XiXj
• di j = distance from P to sidelines XiXj
• Ri j = ratio ai j/di j

Without loss of generality we may assume that P is on the arc X1X2. Then the set of ratios R12, R23, R34, ..., R(n-1)n, Rn1 has an engaging property that, in a sense, extends Theorem 1:

 (2) R12 = R23 + R34 + ... + R(n-1) n + Rn1.

### Proof

The proof is by induction where we start with n = 3, the triangle case. Theorem 1 may serve as the base of the induction.

We proceed with the inductive step. Assume k > 3 that our theorem has been shown for n = k-1. Consider convex k-gon X1X2 ... Xk. This polygon divides the circumcircle into k arcs. We shall assume that P is on the arc X1X2. Consider ΔX1X2X3. This triangle divides circumcircle into three arcs and, by our assumption, P is on the arc X1X2.

X1X3X4 ... Xk is a convex (k-1)-gon and P is on the arc X1X3. (This is because the original k-gon is convex.) Therefore, by the inductive assumption,

R13 = R34 + R45 + ... + Rk1

In ΔX1X2X3,

 R12 = R23 + R13 = R23 + R34 + R45 + ... + Rk1

So the theorem is true for any n ≥ 3.

Bui Quang Tuan also came up with a direct proof that avoids inductive reasoning. Assuming P is on the arc X1X2, consider a triangulation of the polygon X1X2 ... Xn by diagonals emanating from X1. We get triangles X1X2X3, X1X3X4, ..., X1XkX(k+1), ..., X1X(n-1)Xn. Since then P can be considered to lie in arc X1X2, but also in arcs X1X3, X1X4, and so on, Theorem 1, when applied to these triangles, implies the following sequence of identities:

R12 = R23 + R13,
R13 = R34 + R14,
...
R1k = Rk(k + 1) + R1(k + 1),
...
R1(n-2) = R(n-2)(n-1) + R1(n-1),
R1(n-1) = R(n-1)n + R1n.

Summing them up we again obtain the desired result:

R12 = R23 + R34 + ... + R(n-1)n + Rn1.
Copyright © 1996-2018 Alexander Bogomolny

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