Question

In Fig., P is the point on the side BC. Complete each of the following statements using symbol ' = ', '>' or '<' so as to make it true:

(i) AP ... AB + BP
(ii) AP .... AC + PC
(iii) $AP....\frac{1}{2}(AB+AC+BC)$

Answer 1

(i) In triangle APB, AP < AB + BP because the sum of any two sides of a triangle is greater than the third side.

(ii) In triangle APC, AP < AC + PC because the sum of any two sides of a triangle is greater than the third side.

(iii) AP <$\frac{1}{2}\left(\text{AB+AC+BC}\right)$
In triangles ABP and ACP, we can see that:
AP < AB + BP ...(i) (Because the sum of any two sides of a triangle is greater than the third side)
AP < AC + PC ...(ii) (Because the sum of any two sides of a triangle is greater than the third side)

On adding (i) and (ii), we have:

AP + AP < AB + BP + AC + PC
2AP < AB + AC + BC (BC = BP + PC)
AP < $\frac{1}{2}\left(\text{AB+AC+BC}\right)$