Difference between revisions of "Stewart's Theorem"

Revision as of 15:20, 7 January 2020

Statement

Given a triangle $\triangle ABC$ with sides of length $a, b, c$ opposite vertices are $A$ , $B$ , $C$ , 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 $man + dad = bmb + cnc$ , a form which invites mnemonic memorization, i.e. "A man and his dad put a bomb in the sink.")

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

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

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

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

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 $m^2n + n^2m = (m + n)mn = amn$ and $d^2m + d^2n = d^2(m + n) = d^2a$ .

@@ Line 14: / Line 14: @@
 Setting the two left-hand sides equal and clearing [[denominator]]s, we arrive at the equation: <math> c^{2}n + b^{2}m=m^{2}n +n^{2}m + d^{2}m + d^{2}n </math>.
-However, <math>m+n = a</math> so <math>m^2n + n^2m = (m + n)mn = amn and d^2m + d^2n = d^2(m + n) = d^2a</math>.
+However, <math>m+n = a</math> so <math>m^2n + n^2m = (m + n)mn = amn</math> and <math>d^2m + d^2n = d^2(m + n) = d^2a</math>.
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Difference between revisions of "Stewart's Theorem"

Revision as of 15:20, 7 January 2020

Statement

Proof

See also