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>man + dad = bmb + cnc,</math> or Stewart's theorem. | This simplifies our equation to yield <math>man + dad = bmb + cnc,</math> or Stewart's theorem. | ||
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+ | Good Job! You mastered Stewart's Theorem. | ||
== Proof 2 (Pythagorean Theorem) == | == Proof 2 (Pythagorean Theorem) == |
Revision as of 10:42, 24 November 2022
Contents
[hide]Statement
Given a triangle with sides of length and opposite vertices , , , respectively. If cevian is drawn so that , and , we have that . (This is also often written , a phrase which invites mnemonic memorization, i.e. "A man and his dad put a bomb in the sink.")
Proof 1
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.
Good Job! You mastered Stewart's Theorem.
Proof 2 (Pythagorean Theorem)
Let the altitude from to meet at . Let , , and . So, applying Pythagorean Theorem on yields
Since ,
Applying Pythagorean on yields
Substituting , , and in and gives
Notice that
are equal to each other. Thus, Rearranging the equation gives Stewart's Theorem:
~sml1809
Nearly Identical Video Proof with an Example by TheBeautyofMath
~IceMatrix