Question

Prove that the perimeter of a triangle is greater than the sum of its altitudes.

Answer 1

Given : In $\Delta ABC,$

AD, BE and CF are altitudes

To prove : AB + BC + CA > AD + BC + CF

Proof : We know that side opposite to greater angle is greater.

In $\Delta ABD,\angle D={90}^{\circ }$

$\therefore \angle D>\angle B$

$\therefore AB>AD...\left(i\right)$

Similarly, we can prove that

BC > BE and

CA > CF

Adding we get,

AB + BC + CA > AD + BE + CF