Question

Justify the following statements with reasons:

(a) The sum of three sides of a triangle is more than the sum of its altitudes.

(b) The sum of any two sides of a triangle is greater than twice the median drawn to the third side.

(c) Difference of any two sides of triangle is less than the third side.

Answer 1

(a)

Adding (1), (2) and (3):

AD + BE + CF < AB + BC + CA

(b)

Given: ΔABC with median AD

To prove: AB + AC > 2AD

Construction: Produce AD to E such that AD = DE. Join EC.

Proof: In ΔADB and ΔEDC:

AD = DE (Construction)

BD = CD (D is the midpoint of BC)

∠ADB = ∠EDC (Vertically opposite angles)

∴ΔADBΔEDC (SAS congruence criterion)

⇒ AB = EC (CPCT)

In ΔAEC:

AC + CE > AE (Sum of any two sides of a triangles is greater than the third side)

⇒ AC + AB > 2AD (AE = AD + DE = AD + AD = 2AD, EC = AB)

Hence proved

(c) Let ABC be a triangle with sides AB, BC and CA.

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

∴ AB + BC > CA

⇒AB > CA − BC

This shows that the difference of two sides is less than the third side of the triangle.

Similarly, we can prove it for other sides as well.