# Difference between revisions of "Stewart's theorem"

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<cmath>d^2m + d^2n = d^2(m + n) = d^2a.</cmath> | <cmath>d^2m + d^2n = d^2(m + n) = d^2a.</cmath> | ||

This simplifies our equation to yield <math>c^2n + b^2m = amn + d^2a,</math> or Stewart's theorem. | This simplifies our equation to yield <math>c^2n + b^2m = amn + d^2a,</math> or Stewart's theorem. | ||

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+ | ==Nearly Identical Video Proof with an Example by TheBeautyofMath== | ||

+ | https://youtu.be/jEVMgWKQIW8 | ||

+ | |||

+ | ~IceMatrix | ||

== See also == | == See also == |

## Revision as of 15:15, 11 September 2021

## Contents

## Statement

Given a triangle with sides of length opposite vertices are , , , respectively. If cevian is drawn so that , and , we have that . (This is also often written , 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 at angle and in triangle at angle , we get the equations

Because angles and are supplementary, . We can therefore solve both equations for the cosine term. Using the trigonometric identity gives us

Setting the two left-hand sides equal and clearing denominators, we arrive at the equation: . However, so and This simplifies our equation to yield or Stewart's theorem.

## Nearly Identical Video Proof with an Example by TheBeautyofMath

~IceMatrix