×

# 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
7 months, 3 weeks ago

Sort by:

Check out stewart's theorem. Staff · 7 months, 2 weeks 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).

$\large Q. E. D.$ · 7 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) · 7 months, 2 weeks ago

Notice that $$AD=c \cos A$$ and $$CD=a \cos C$$. · 7 months, 2 weeks ago