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 ABCABC I draw the height BDBD and I call WW the value W=AB2+BC2AC22W=\frac{AB^2+BC^2-AC^2}{2} (that is equal to ABBCcos(B^)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) BC2=DCAC+WBC^2=DC*AC+W and on the other side AB2=ADAC+WAB^2=AD*AC+W but I can't explain myself these equalities, could you help me please? Thanks :)

Note by Matteo Monzali
3 years, 5 months ago

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Check out stewart's theorem.

Calvin Lin Staff - 3 years, 5 months ago

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Let's put AB=cBC=aCA=bAB=c \\ BC=a \\ CA=b

Then the conjectured equations are equivalent to the following equations: a=bcosC+ccosBc=acosB+bcosAa=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).

Q.E.D. \large Q. E. D.

Deeparaj Bhat - 3 years, 5 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 - 3 years, 5 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 - 3 years, 5 months ago

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Notice that AD=ccosAAD=c \cos A and CD=acosCCD=a \cos C.

Deeparaj Bhat - 3 years, 5 months ago

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