Prove that the figure obtained by joining the mid-points of the adjacent sides of a rectangle is a rhombus.

Let ABCD be the rectangle and P, Q, R and S be the mid-points of sides AB, BC, CD and DA respectively. Join PQRS.

By mid-point theorem,

The line segment joining the mid-points of any two sides of a triangle is parallel to the third side and is equal to half of it.

Join AC and BD.

We know that,

Diagonals of a rectangle are equal.

∴ AC = BD = x (let)

In Δ ABC, P and Q are the mid-points of sides AB and BC respectively.

∴ PQ || AC and PQ = $\dfrac{1}{2}AC = \dfrac{1}{2}x$ (By mid-point theorem) ……..(1)

In Δ ADC, S and R are mid-points of sides AD and CD respectively.

∴ SR || AC and SR = $\dfrac{1}{2}AC = \dfrac{1}{2}x$ (By mid-point theorem) ………(2)

From equations (1) and (2), we get :

PQ || SR and PQ = SR

In Δ ABD, P and S are the mid-points of sides AB and AD respectively.

∴ PS || BD and PS = $\dfrac{1}{2}BD = \dfrac{1}{2}x$ (By mid-point theorem) ……..(3)

In Δ BDC, Q and R are mid-points of sides BC and CD respectively.

∴ QR || BD and QR = $\dfrac{1}{2}BD = \dfrac{1}{2}x$ (By mid-point theorem) ………(4)

From equations (3) and (4), we get :

PQ || SR and PS = QR

By using equation (1), (2), (3) and (4), we get :

⇒ PQ = QR = SR = PS

Since, opposite sides are parallel and all the sides are equal.

∴ PQRS is a rhombus.

Hence, proved that the figure obtained by joining the mid-points of the adjacent sides of a rectangle is a rhombus.