Question

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

Answer 1

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

Draw $AD,BE,CF$ perpendicular from $A,B,C$ to opposite sides meeting sides $BC,AC,AB$ at $D,E,F$ respectively.

Now perpendicular $A{D}^2=A{C}^2-C{D}^2$

$\Rightarrow A{D}^2<A{C}^2$

$\Rightarrow AD<AC$

$\Rightarrow AD<b$ .... (1)

Also, $B{E}^2=A{B}^2-A{E}^2$

$\Rightarrow B{E}^2<A{B}^2$

$\Rightarrow BE<AB$

$\Rightarrow BE<c$ ..... (2)

Likewise $C{F}^2=B{C}^2-B{F}^2$

$\Rightarrow C{F}^2<B{C}^2$

$\Rightarrow CF<BC$

$\Rightarrow CF<a$ ..... (3)

Adding inequalities, (1), (2) and (3), we get

$AD+BE+CF<a+b+c$

Hence Proved.