Question

Prove that any two sides of a triangle are together greater than twice the median drawn to the third side. [4 MARKS]

Answer 1

Concept : 1 Mark
Process : 2 Marks
Proof : 1 Mark

Given : A $△$ABC in which AD is a median.

To prove: AB + AC $>$ 2 AD

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

Proof:

In $△$s ADB and EDC, we have

AD = DE [By construction]

BD = DC [$\because$ D is the midpoint of BC]

$\angle$ADB = $\angle$EDC [Vertically opp. angles]

$\therefore △$ADB $\cong$ $△$ EDC [by SAS criterion of congruence]

$\Rightarrow$ AB = EC [Corresponding parts of congruent triangles are equal]

Now in $△$AEC, we have

AC + EC $>$ AE [$\because$ Sum of any two sides of a $△$ is greater than the third]

$\Rightarrow$ AC + AB $>$ 2 AD

[$\because$ AD = DE.$\therefore$AE = AD + DE = 2AD and EC = AB]

Hence proved.