An acute-angled triangle

Question

If $a$, $b$ and $c$ are the sides of the triangle $ABC$, then prove that the triangle with sides $\sqrt{a^2+(b-c)^2}$, $\sqrt{b^2+(a-c)^2}$, $\sqrt{c^2+(a-b)^2}$ is acute.

Answer 1

To prove a triangle acute, it's enough to prove that cosine of all the angles of the triangle is positive.

Let $\rm A , B,C$ be the sides of triangle.

$\rm A = \sqrt{a^2+(b-c)^2}$

$\rm B= \sqrt{ b^2+(a-b)^2}$

$\rm C= \sqrt{c^2+(a-b)^2}$

$$\cos \rm A= \frac{B^2+C^2-A^2}{2BC}$$

$$\cos \rm B= \frac{A^2+C^2-B^2}{2AC}$$ $$\cos \rm C= \frac{A^2+B^2-C^2}{2AB}$$

Plugging in the values we get $$\cos \rm A= \frac{a^2+b^2+c^2-2ab+2bc-2ac}{2BC} = \frac{(b+c-a)^2}{2BC} >0$$

$$\cos \rm B= \frac{a^2+b^2+c^2-2ab-2bc+2ac}{2BC} = \frac{(a+c-b)^2}{2BC} >0$$

$$\cos \rm C= \frac{a^2+b^2+c^2+2ab-2bc-2ac}{2BC} = \frac{(a+b-c)^2}{2BC} >0$$

Answer 2

A triangle with sides a,b,and c (in ascending order) is acute if $a^2+b^2>c^2$, obtuse if $a^2+b^2<c^2$.