Proof of Stewart’s theorem from Ptolemy’s theorem

by SomeonecoolLovesMaths, Mar 17, 2024, 6:19 PM

Stewart’s theorem: If $D$ is a point on side $BC$ of $\triangle ABC$ such that $BD=m, DC=n, AD=p$ then, $b^2 m+c^2 n=a(p^2+mn)$ , where $a,b,c$ are sides of $\triangle ABC$ , as usual.

https://cdn.aops.com/images/e/7/8/e7831255cfca5a657449b9f01158d51686ab248e.png

Proof: The $2$ red triangles in the figure are similar triangles and so are the $2$ blue triangles. Hence, $\frac{PB}{b}=\frac{m}{p}=\frac{PD}{n} and \frac{PC}{c}=\frac{n}{p}$ .
Now, by Ptolemy’s theorem in cyclic quadrilateral $ABPC$ , we have, $AC\cdot PB+AB\cdot PC=BC\cdot AP$ .
Hence, $b\cdot\frac{bm}{p}+c\cdot\frac{cn}{p}=a(p+\frac{mn}{p}) or, b^2m+c^2n=a(p^2+mn)$ .(Proved)

geometry

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Proof of Stewart’s theorem from Ptolemy’s theorem

by SomeonecoolLovesMaths, Mar 17, 2024, 6:19 PM

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