Question

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

Answer 1

Given : In $\Delta ABC,AD,BEandCF\text{are the altitude of}\Delta ABC$

To prove : AD + BE + CF < AB + BC + CA

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

Then other two angles are acute

$\because \angle B<\angle D$

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

Similarly, in $\Delta BECand\Delta ABE$ we can prove

$thatBE<BC...\left(ii\right)$

$andCF<CA...\left(iii\right)$

Adding (i), (ii), (iii)

$AD+BE+CF<AB+BC+CA$