Question

17. Prove that in a triangle the sum of the medians is less than the perimeter.
18. From the figure, prove that PA+AQ <PB+BQ, where ZPAX= 4QAY​

Answer 1

Let ABC be the triangle and D,E and F are midpoint of BC, CA and AB respectively.

Have to call that the sum of two sides of a triangle is greater than twice the median bisecting the third side.

Hence in ΔABD AD is median

⇒AB+AC>2(AD)

Similarly, we get

BC+AC>2CF

BC+AB>2BE

On adding the above in equations, we get

(AB+AC)+(BC+AC)+(BC+AB)>2AD+2CD+2DE

⇒2(AB+BC+AC)>2(AD+BE+CF)

∴ AB+BC+AC>AD+BE+CF

solution