AB and CD are respectively the smallest and longest sides of a quadrilateral ABCD (see the given figure). Show that ∠A > ∠C and ∠B > ∠D.

Solution:



Let us join AC.

$\ln \Delta \mathrm{ABC}$,

$\mathrm{AB}<\mathrm{BC}(\mathrm{AB}$ is the smallest side of quadrilateral $\mathrm{ABCD})$

$\therefore \angle 2<\angle 1$ (Angle opposite to the smaller side is smaller) ... (1)

In $\triangle \mathrm{ADC}$,

AD < CD (CD is the largest side of quadrilateral ABCD)

$\therefore \angle 4<\angle 3$ (Angle opposite to the smaller side is smaller)..(2)

On adding equations (1) and (2), we obtain

$\angle 2+\angle 4<\angle 1+\angle 3$

$\Rightarrow \angle C<\angle A$

$\Rightarrow \angle \mathrm{A}>\angle \mathrm{C}$

Let us join BD.



In $\triangle \mathrm{ABD}$,

$\mathrm{AB}<\mathrm{AD}$ ( $\mathrm{AB}$ is the smallest side of quadrilateral $\mathrm{ABCD}$ )

$\therefore \angle 8<\angle 5$ (Angle opposite to the smaller side is smaller) ... (3)

In $\triangle \mathrm{BDC}$,

$\mathrm{BC}<\mathrm{CD}$ ( $\mathrm{CD}$ is the largest side of quadrilateral $\mathrm{ABCD}$ )

$\therefore \angle 7<\angle 6$ (Angle opposite to the smaller side is smaller) ... (4)

On adding equations (3) and (4), we obtain

$\angle 8+\angle 7<\angle 5+\angle 6$

$\Rightarrow \angle \mathrm{D}<\angle \mathrm{B}$ $\Rightarrow \angle B>\angle D$