Difference between revisions of "Stewart's Theorem"
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Now we set the two equal and arrive at Stewart's theorem: <math> c^{2}n + b^{2}m=m^{2}n +n^{2}m + d^{2}m + d^{2}n </math>. | Now we set the two equal and arrive at Stewart's theorem: <math> c^{2}n + b^{2}m=m^{2}n +n^{2}m + d^{2}m + d^{2}n </math>. | ||
| + | However, since <math>m+n</math> can be written as a, we get the more common form: <math>cnc+bmb=man+dad</math> | ||
== See also == | == See also == | ||
Revision as of 07:51, 5 November 2006
Statement
If a cevian of length d is drawn and divides side a into segments m and n, then
Proof
For this proof, we will use the law of cosines and the identity 
.
Label the triangle 
with a cevian extending from 
onto 
, label that point 
. Let CA = n. Let DB = m. Let AD = d. We can write two equations:
When we write everything in terms of cos(CDA) we have:
Now we set the two equal and arrive at Stewart's theorem: 
.
However, since 
can be written as a, we get the more common form: 
