Question

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

Answer 1

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

Consider the attached figure while going through the following steps.

Let ABC be a triangle with sides AB, BC and CA

Let the length of sides be a, b and c.

Let the altitudes be AD, BE and CF.

Now, we have,

AD^2 = AC^2 - CD^2

⇒ AD^2  < AC^2

⇒ AD < AC

∴ AD < b .............(1)

Now, we have,

BE^2 = AB^2 - AE^2

⇒ BE^2  < AB^2

⇒ BE < AB

∴ BE < c .............(2)

Now, we have,

CF^2 = BC^2 - BF^2

⇒ CF^2  < BC^2

⇒ CF < BC

∴ CF < a .............(3)

Adding (1), (2) and (3), we have,

(AD + BE + CF) < (b + c + a)

where, a, b and c are the lengths of sides of a triangle.

AD, BE and CF are the altitudes.

Hence, it is proved that, the sum of three altitudes of a triangle is less than the sum of its sides.