Question

In the figure, $D$ is a point on $BC$. Prove that: $AB+BC+CA>2AD$.

Answer 1

Given: $△ABC$ in which $D$ is a point on $BC$

To prove: $AB+BC+CA>2AD$

Construction: $AD$ is joined.

Proof: In $△ABD,AB+BD>AD$ [$\because$, the sum of any

two sides of a triangle isalways greater than the third side] ---$\left(1\right)$

In $△ADC,AC+DC>AD$ [$\because$, the sum of any two sides of a triangle is always greater than the third side] ---- $\left(2\right)$

Adding $\left(1\right)$ and $\left(2\right)$ we get,

$AB+BD+AC+DC>AD+AD$

$\Rightarrow AB+(BD+DC)+AC>2AD$

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

Hence proved