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 :)

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TopNewestCheck out stewart's theorem. – Calvin Lin Staff · 9 months, 2 weeks 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 · 9 months, 3 weeks ago

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– Matteo Monzali · 9 months, 3 weeks ago

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)Log in to reply

– Deeparaj Bhat · 9 months, 3 weeks ago

Notice that \(AD=c \cos A\) and \(CD=a \cos C\).Log in to reply

– Ameya Daigavane · 9 months, 3 weeks ago

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.Log in to reply