A strange connection in triangles?

Casually I found a relation between the sides of any triangle but I don't know if it is a thing that you can obtain from other theorems or no

Given a triangle \(ABC\) I draw the height \(BD\) and I call \(W\) the value \(W=\frac{AB^2+BC^2-AC^2}{2}\) (that is equal to \(AB*BC*\cos(\hat{B})\) because of the law of cosines)

Then you can find (I've found this casually, but you can verify easily thanks to GeoGebra or similar programs) \(BC^2=DC*AC+W\) and on the other side \(AB^2=AD*AC+W\) but I can't explain myself these equalities, could you help me please? Thanks :)

Note by Matteo Monzali
2 years, 4 months ago

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Check out stewart's theorem.

Calvin Lin Staff - 2 years, 4 months ago

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Let's put \[AB=c \\ BC=a \\ CA=b\]

Then the conjectured equations are equivalent to the following equations: \[a=b\cos C + c\cos B \\ c=a\cos B + b\cos A\]

But these equivalent equations are true for any arbitrary triangles (try proving this).

\[ \large Q. E. D. \]

Deeparaj Bhat - 2 years, 4 months ago

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Yes I already know these equations, but I don't understand how they can be equivalent to the ones I wrote (I tried to put them equal but I can't reach a result)

Matteo Monzali - 2 years, 4 months ago

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Notice that \(AD=c \cos A\) and \(CD=a \cos C\).

Deeparaj Bhat - 2 years, 4 months ago

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Just as a hint, try drawing an altitude from one of the vertices to a side, and then use the two right triangles formed on either side.

Ameya Daigavane - 2 years, 4 months ago

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