Line segments ab and cd intersect each other at point o. prove that the sum of the distances between any point on the plane to points a, b, c, and d is not lesser than oa+ob+oc+od.

Solution: Two line segments AB and CD of length m and n intersect each other at point O.

Considering it is a two dimensional plane and O being the origin, and AB being X axis and CD being Y axis.

Coordinates of A=(m/2,0), Coordinate of B=(-m/2,0)

Coordinate of C=(0,n/2), Coordinate of D=(0,-n/2)

Let P(m/2, n/2) be any point in that plane.

OA+OB+OC+OD=m/2+m/2+n/2+n/2=m+n----------------(1)

PA+PB+PC+PD= n/2++m/2+

As, is always greater than m and   is always greater than n .

So,PA+PB+PC+PD will always be>m+n.      ----------(2)

Combining (1) and (2), we arrived at the result

PA+PB+PC+PD>OA+OB+OC+OD

step-by-step explanation:

ddd

the answer is 510

step-by-step explanation:

15 • 13 = 195

15 • 12 = 180

15 • 5 = 75

12 • 5 = 60

195 + 180 = 375

375 + 75 = 450

450 + 60 = 510