Stewart's Theorem

 Stewart's theorem





Proof:
Drop a perpendicular from A to BC at M.
S.t. BM + MD = BD and CM - MD = CD

AB^2 =  AM^2 + BM^2
=> AM^2 = AB^2 - BM^2 = c^2 - (BD - MD)^2 = c^2 - (m - MD^2)
Similarly:
AM^2 = AC^2 - MC^2 = b^2 - (n - MD)^2
Now equate both to eliminate MD^2.
Then multiply first equation by 'n', second by 'm' to eliminate the MD term.

You will get the proof.

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