Question

In a quadrilateral ABCD, P is an internal point. Prove that AP + BP + CP + DP > Semi-perimeter of quadrilateral ABCD.

Answer 1

ABCD is a quadrilateral, P is an internal point.

We know that, The sum of two sides of a triangle is greater than the third side.

In ΔDPC, DP + CP > DC    ...(1)

In ΔAPB, AP + BP > AB    ...(2)

In ΔCPB, CP + BP > CB    ...(3)

In ΔAPD, DP + AP > AD    ...(4)

Adding equation (1), (2), (3) and (4),

DP + CP + AP + BP + CP + BP + DP + AP > DC + AB + CB + AD

2(AP + BP + CP + DP) > AB + CB + DC + AD

2(AP + BP + CP + DP) > Semi-perimeter of quadrilateral ABCD

Hence, proved.