Question

Prove that the sum of three altitudes of a triangle is less than the sum of three sides of the triangle.

Answer 1

Let $ABC$ be the $△$ with sides $BC,AC$ and $AB$ opposite to the angles $A,B$ and $C$ being in length equal to $a,b$ and $c$ respectively.

To prove:- $AL+BM+CN<AB+BC+CA$

Construction:- Draw perpendiculars $AL,BM$ and $CN$ from $A,B$ and $C$ to opposite sides meeting respectively.

Now,

In $△ACL$,

Using pythagoras theorem,

${AC}^2={AL}^2+{CL}^2$

$\Rightarrow {AC}^2>{AL}^2$

$\Rightarrow AC>AL.....\left(1\right)$

Similarly,

$AB>BM.....\left(2\right)$

$BC>CN.....\left(3\right)$

Adding equation $\left(1\right),\left(2\right)\&\left(3\right)$, we get]

$AB+BC+AC>AL+BM+CN$

Hence proved.