In Fig., P is the point on the side BC. Complete each of the following

1.	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…. ½ (AB + AC + BC)
Answer» (i) In △APB, AP < AB + BP because the sum of any two sides of a triangle is greater than the third side. (ii) In △APC, AP < AC + PC because the sum of any two sides of a triangle is greater than the third side. (iii) AP < ½ (AB + AC + BC) In △ABP and △ACP, we can write as 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 < (AB – AC + BC)

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…. ½ (AB + AC + BC)

Answer»

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

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

(iii) AP < ½ (AB + AC + BC)

In △ABP and △ACP, we can write as

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 < (AB – AC + BC)