# 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
3 years, 7 months ago

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Check out stewart's theorem.

Staff - 3 years, 7 months ago

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).

$\large Q. E. D.$

- 3 years, 7 months 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.

- 3 years, 7 months 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)

- 3 years, 7 months ago

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

- 3 years, 7 months ago