proof of Stewart’s theorem


Let θ be the angle AXB.

Cosines law on AXB says c2=m2+p2-2pmcosθ and thus

cosθ=m2+p2-c22pm

Using cosines law on AXC and noting that ψ=AXC=180-θ and thus cosθ=-cosψ we get

cosθ=b2-n2-p22pn.

From the expressions above we obtain

2pn(m2+p2-c2)=2pm(b2-n2-p2).

By cancelling 2p on both sides and collecting we are led to

m2n+mn2+p2n+p2m=b2m+c2n

and from there mn(m+n)+p2(m+n)=b2m+c2n. Finally, we note that a=m+n so we conclude that

a(mn+p2)=b2m+c2n.

QED

Title proof of Stewart’s theorem
Canonical name ProofOfStewartsTheorem
Date of creation 2013-03-22 12:38:37
Last modified on 2013-03-22 12:38:37
Owner Mathprof (13753)
Last modified by Mathprof (13753)
Numerical id 7
Author Mathprof (13753)
Entry type Proof
Classification msc 51-00
Related topic StewartsTheorem
Related topic ApolloniusTheorem
Related topic CosinesLaw
Related topic ProofOfApolloniusTheorem2