Stewart's Theorem
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, $$ (Error compiling LaTeX. Unknown error_msg)m+n = am^2n + n^2m = (m + n)mn = amnd^2m + d^2n = d^2(m + n) = d^2ac^2n + b^2m = amn + d^2a,$$ (Error compiling LaTeX. Unknown error_msg) or Stewart's Theorem.