Difference between revisions of "Stewart's Theorem"
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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>. | 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</math> | + | However, <math>m+n = a</math> so <math>m^2n + n^2m = (m + n)mn</math>. |
== See also == | == See also == |
Revision as of 00:45, 13 July 2019
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 .