Question

ABCD
is a quadrilateral. M is the midpoint of
diagonal AC and N is the midpoint of diagonal BD.
Prove that :
AB+ BC? + CD + DAP = AC? + BD+ 4MN?.​

Answer 1

In quadrilateral ABCD , draw the diagonals AC and BD which meet at point P.

we know that sum of any two sides of a triangle is always greater than the third

side. In ∆PAB. , PA+PB > AB………………..(1)

In ∆ PBC. , PB+PC > BC………………………..(2).

In. ∆ PCA. , PC+PA > CA……………………….(3).

In ∆ PAD. , PA +PD > DA………………………(4).

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

2.(PA+PB+PC+PD) > AB+BC+CD+DA..

or. 2{(PA+PC)+(PB+PD)} > AB+BC+CD+DA.

or. 2.{ AC+BD} > AB+BC+CD+DA. Proved.