Question

ABCD is a parallelogram in which P, Q, R, S are mid point of sides AB, BC, CO and AD. show that-:PB = RS​

Answer 1

Answer:

Let us join AC and BD.

In ΔABC,

P and Q are the mid-points of AB and BC respectively.

∴ PQ || AC and PQ = AC (Mid-point theorem) ... (1)

Similarly in ΔADC,

SR || AC and SR = AC (Mid-point theorem) ... (2)

Clearly, PQ || SR and PQ = SR

Since in quadrilateral PQRS, one pair of opposite sides is equal and parallel to  

each other, it is a parallelogram.

∴ PS || QR and PS = QR (Opposite sides of parallelogram)... (3)

In ΔBCD, Q and R are the mid-points of side BC and CD respectively.

∴ QR || BD and QR =BD (Mid-point theorem) ... (4)

However, the diagonals of a rectangle are equal.

∴ AC = BD …(5)

By using equation (1), (2), (3), (4), and (5), we obtain

PQ = QR = SR = PS

Therefore, PQRS is a rhombus

Answer 2

Answer:

as its alternate interior angles

angle pbq = angle rsp