# Stewart's theorem

In geometry, Stewart's theorem yields a relation between a lengths of the sides of the triangle and the length of a cevian of the triangle. Its name is in honor of the Scottish mathematician Matthew Stewart who published the theorem in 1746.[1]

## Theorem

Let $a$, $b$, and $c$ be the lengths of the sides of a triangle. Let $d$ be the length of a cevian to the side of length $a$. If the cevian divides $a$ into two segments of length $m$ and $n$, with $m$ adjacent to $c$ and $n$ adjacent to $b$, then Stewart's theorem states that

$b^2m + c^2n = a(d^2 + mn).\,$

Apollonius' theorem is the special case where d is the length of the Median.

The theorem may be written somewhat more symmetrically using signed lengths of segments, in other words the length AB is taken to be positive or negative according to whether A is to the left or right of B in some fixed orientation of the line. In this formulation, the theorem states that if A, B, and C are collinear points, and P is any point, then[2]

$PA^2\cdot BC+PB^2\cdot CA-PC^2\cdot AB - BC\cdot CA\cdot AB =0.\,$

## Proof

The theorem can be proved as an application of the law of cosines:[3]

Let θ be the angle between m and d and θ′ the angle between n and d. Then θ′ is the supplement of θ and cos θ′ = −cos θ. The law of cosines for θ and θ′ states \begin{align} c^2 &= m^2 + d^2 - 2dm\cos\theta \\ b^2 &= n^2 + d^2 - 2dn\cos\theta' \\ &= n^2 + d^2 + 2dn\cos\theta.\, \end{align}

Multiply the first equation by n, the second equation by m, and add to eliminate cos θ, obtaining \begin{align} &b^2m + c^2n \\ &= nm^2 + n^2m + (m+n)d^2 \\ &= (m+n)(mn + d^2) \\ &= a(mn + d^2), \\ \end{align} which is the required equation.

Alternatively, the theorem can be proved by drawing a perpendicular from the vertex of the triangle to the base and using the Pythagorean theorem to write the distances b, c, and d in terms of the altitude. The left and right hand sides of the equation then reduce algebraically to the same expression.[4]