Adjacent sides of a parallelogram are equal and one of diagonals is equal to any one of the sides of this parallelogram. Show that its diagonals are in the ratio $\sqrt{3} : 1$.

Let ABCD be the required parallelogram.

∴ AB = CD and BC = AD. (Opposite sides of parallelogram are equal)

Given,

Adjacent sides of a parallelogram are equal.

∴ AB = BC.

∴ AB = BC = CD = AD

Since, all sides of parallelogram are equal.

∴ ABCD is a rhombus.

Given, one of the diagonals is equal to its sides. Let diagonal BD be equal to sides.

∴ AB = BC = CD = AD = BD = a (let).

From figure,

⇒ BO = $\dfrac{BD}{2} = \dfrac{a}{2}$ (Since, in a rhombus diagonals bisect each other at right angle).

Hence, △ AOB is right-angled at O.

In △ AOB,

By pythagoras theorem,

⇒ AB² = BO² + AO²

⇒ a² = $\Big(\dfrac{a}{2}\Big)^2$ + AO²

⇒ AO² = $a^2 - \dfrac{a^2}{4}$

⇒ AO² = $\dfrac{4a^2 - a^2}{4}$

⇒ AO² = $\dfrac{3a^2}{4}$

⇒ AO = $\dfrac{\sqrt{3}a}{2}$,

⇒ AC = 2AO = $2 \times \dfrac{\sqrt{3}a}{2} = \sqrt{3}a$ units.

The ratio of the diagonals is:

$\Rightarrow \dfrac{AC}{BD} = \dfrac{\sqrt{3}a}{a} = \dfrac{\sqrt{3}}{1}$

∴ AC : BD = $\sqrt{3}$ : 1.

Hence, proved that diagonals are in the ratio $\sqrt{3}$ : 1.