Stewart

Stewart's Theorem

Consider a Cevian AD from vertex A of the triangle ABC, intersecting the opposite side BC in ratio BD/DC = r/s. Then the length AD² is:

The proof results by applying the cosine theorem on triangles ABD, ADC.
A question that may arise is if there exists a ratio r/s, such that all corresponding segments AD, BE, CF are equal.
Setting this common length AD² = BE² = CF² = d², we get the equation

Considering this as a linear system in r², s², rs, the determinant of the matrix is W = (a⁶+b⁶+c⁶-3(abc)²). W = 0 when the triangle is equilateral. Then, for every (r,s) we find a solution. When W is nonzero, then the solution is of the form (r², s², rs) = m(2, 2, -1), which gives an incompatible system for r,s.
Two instances of the formula are well known: Applying this formula for the median (r=s=1), gives:
m_a² = (2(b²+c²)-a²)/4.
And for the internal bisector of the angle A (r = c, s = b):
u² = bc(1- (a /(b+c))²), whereas the external bisector satisfies:
v² = bc((a /(b-c))²-1).
For points D = (1-t)B+tC, BD/DC = t/(1-t) = r/s with r+s=1. The above formula becomes:
AD² = tb²+(1-t)c²-t(1-t)a². This can be used to determine the (at most) two values of t, for which the cevian AD has a given length. In particular the minimum value, corresponding to the altitude from A, occurss when 2ta² + (b²-a²-c²) = 0, and the expresion for the ratio is:
BD/DC = t/(1-t) = (a²+c²-b²)/(a²+b²-c²).
For a fixed t, the corresponding lengths AD², BE², CF² satisfy the equations:

Eliminating t from the equations we find two equations determining AD², BE², CF², once one of the is known.
See the file SquaresCombination.html for a nice application of the theorem of Stewart.

References

Steiner, J. Werke Bd. I New York, Chelsea, 1971, pp. 15.

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Stewart's Theorem

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References