the diagonals PR and QS of a rhombus intersect each other at point O. prove that 2(PQ) + 2(QR) + 2(RS)+ 2(PS) = 4 (2(OP)+2(OQ))

It is proved that 2PQ + 2SR + 2QR + 2 PS = 4(2(OP)+2(OQ))

Step-by-step explanation:

In a rhombus, all sides are equal.

Thus, in this question;

PQ = SR = QR = PS

By inspection, QS is the same dimension as the four sides. So, QS = PQ

Thus, OQ = PQ/2

For OP, we can find it using Pythagoreas theorem since the angle that divides the diagonals is 90°

Thus;

|OP|² + |OQ|² = |PQ|²

Earlier, we saw that;OQ = PQ/2

Thus;

|OP|² + |PQ/2|² = |PQ|²

|OP|² = |PQ|² - |PQ/2|²

|OP|² = |PQ/2|²

Taking square root of both sides, we have;

OP = PQ/2

So,going back to the question, on the right hand side, we have;

4(2(OP) + 2(OQ))

Let's put,

PQ/2 for OQ and OP as gotten earlier

So,

4(2(PQ/2) + 2(PQ/2)) = 4(PQ + PQ)

Since PQ = SR = QR = PS, we can rewrite as;

4PQ + 4PQ = (PQ + SR + QR + PS) + (PQ + SR + QR + PS) = 2PQ + 2SR + 2QR + 2 PS

This is equal to the left hand side, so the equation is proved correct.

∡ eia and ∡ kij are congruent

step-by-step explanation: