Question

Show that the sum of three altitudes ora triangle is less than the sum of three
sides of the triangle​

Answer 1

Correct Question:-

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

AnswEr:-

In ∆ABC, AD, BE & CF are the median to the sides BC, AB & AD respectively.

We know that the sum of the any two sides of the triangle is greater than twice the median bisecting the third side.

$\rule{150}3$

Here,

In ∆ABD

∠ D = 90° & ∠B is acute.

∠ D > ∠B

∴ AB > AD ______eq(1)

[Sides Opposite to the greater anglem is longer]

In ∆ACD

∠ D = 90° and ∠C is acute.

∠ D = ∠C

∴ AC > AD _______eq (2)

Adding eqn (1) & (2) we get,

AB + AC > 2AD ________eq(3)

Same as, we can prove that

BC + BA > 2BE _________eq(4)

∴ BC = AB

And CA + CB > 2CF ________eq(5)

∴ CF = AB

$\rule{150}3$

Now, Adding eqn (3),(4) & (5)

↬ 2(AB + BC + CA) > 2(AD + BE + CF)

↬ AB + BC + CA > AD + BE + CF

↬AD + BE + CF < AB + BC + CA

Hence Proved!

$\rule{150}3$