Example 3 is very confusing. Please tell in easy way

 

Dear Student!

In the given figure BD ⊥ AC.

It is known that the sum of any two sides of a triangle is always greater than the third side.

In ΔABD, DA + AB > BD ...(1)

Similarly In ΔBCD, BC + CD > BD ...(2)

On adding (1) and (2), we obtain

(DA + AB) +  (BC + CD) > BD + BD

⇒ (DA + CD) + (AB + BC) > 2BD  ...(3)

It can be observed from the figure that DA + CD = CA. On putting this value in equation (3), we obtain

CA + AB + BC > 2BD 

⇒ AB + BC + CA > 2BD

or 2BD < AB + BC + CA

Hope it helps.

Best wishes!