Question

Al. Diagonals PR and QS of a rhombus intersect
each other at point O. Prove that
PQP + QR2 + RS? + PS2 = 4(OP? + OQ?).​

Answer 1

Answer:

In a rhombus, all sides are equal.  That is, PQ = QR = RS = PS, so

   PQ² + QR² + RS² + PS² = 4PQ².

In a rhombus, the diagonals meet at right angles, so ΔOPQ is a right angled triangle with hypotenuse PQ.  So by Pythagoras' Theorem

   PQ² = OP² + OQ².

Putting these together gives

   PQ² + QR² + RS² + PS² = 4PQ² = 4 ( OP² + OQ² ).

Hope this helps!