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Stewart's Theorem

From AoPSWiki

Statement

Given a triangle \triangle ABC with sides of length a, b, c opposite vertices A, B, C, respectively. If cevian AD is drawn so that BD = m, DC = n and AD = d, we have that b^2m + c^2n = amn + d^2a. (This is also often written cnc + bmb = man + dad, a form which invites mnemonic memorization, e.g. "A man and his dad put a bomb in the sink.")

Image:Stewart's_theorem.png

Proof

Applying the Law of Cosines in triangle \triangle ABD at angle \angle ADB and in triangle \triangle ACD at angle \angle CDA, we get the equations

  • n^{2} + d^{2} - 2nd\cos{\angle CDA} = b^{2}
  • m^{2} + d^{2} - 2md\cos{\angle ADB} = c^{2}

Because angles \angle ADB and \angle CDA are supplementary, m\angle ADB = 180^\circ - m\angle CDA. We can therefore solve both equations for the cosine term. Using the trigonometric identity \cos{\theta} = -\cos{(180^\circ - \theta)} gives us

  • \frac{n^2 + d^2 - b^2}{2nd} = \cos{\angle CDA}
  • \frac{c^2 - m^2 -d^2}{2md} = \cos{\angle CDA}

Setting the two left-hand sides equal and clearing denominators, we arrive at the equation: c^{2}n + b^{2}m=m^{2}n +n^{2}m + d^{2}m + d^{2}n. However, m+n = a so m^2n + n^2m = (m + n)mn and we can rewrite this as c^2n + b^2m = amn + d^2a.

See also

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Want to learn how to tackle those tough MATHCOUNTS and AMC counting and probability problems? Check out Art of Problem Solving's Introduction to Counting & Probability by David Patrick.
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