If the angles of a quadrilateral, taken in order, are in the ratio 1 : 2 : 3 : 4, prove that it is a trapezium.

Let a trapezium be ABCD.

Let angles be x, 2x, 3x and 4x taken in order.

Sum of angles of quadrilateral = 360°.

∴ x + 2x + 3x + 4x = 360°

⇒ 10x = 360°

⇒ x = $\dfrac{360°}{10}$

⇒ x = 36°.

⇒ ∠A = x = 36°,

⇒ ∠B = 2x = 2(36°) = 72°

⇒ ∠C = 3x = 3(36°) = 108°

⇒ ∠D = 2x = 4(36°) = 144°.

From figure,

∠A and ∠D are co-interior angles and their sum = 180° (36° + 144°).

Sum of co-interior angles = 180°, which means AB and CD are parallel.

Also,

Opposite angles sum i.e.

∠A + ∠C = 36° + 108° = 144°,

∠B + ∠D = 72° + 144° = 216°.

Since, the sum of opposite angles ≠ 180° (Property of trapezium).

Hence, proved that ABCD is a trapezium.