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.") That is Stewart's Theorem. I know, it's easy to memorize.
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
Applying Pythagorean on yields
Substituting , , and in and gives
are equal to each other. Thus, Rearranging the equation gives Stewart's Theorem:
Proof 3 (Barycentrics)
Let the following points have the following coordinates:
Our displacement vector has coordinates . Plugging this into the barycentric distance formula, we obtain Multiplying by , we get . Substituting with , we find Stewart's Theorem:
Nearly Identical Video Proof with an Example by TheBeautyofMath