Question

$AB$ and $CD$ are respectively the smallest and largest side of $\square ABCD$, Prove that$\angle A>\angle C$ and $\angle B>\angle D$.

Answer 1

$AB$ is the smallest side ----- ( 1 )

$CD$ is the longest side ----- ( 2 )

Now, join $A$ and $C$ and mark the angles.

In $△$ABC,

$\Rightarrow$ $BC>AB$ [ From ( 1 ) ]

$\therefore$ $\angle 1>\angle 2$ [ Angle opposite to the longer side is greater ]

In $△ADC,$

$\Rightarrow$ $CD>AD$ [ From ( 2 ) ]

$\therefore$ $\angle 3>\angle 4$ [ Angle opposite to the longer side is greater ]

Adding equations

$\Rightarrow$ $\angle 1+\angle 3>\angle 2+\angle 4$

$\Rightarrow$ $\angle A>\angle C$ ---- Hence proved

Now, solving for $\angle B>\angle D$

Join $B$ and $D,$ and mark the angles

In $△ABD,$

$\Rightarrow$ $AD>AB$ [ From ( 1 )]

$\therefore$ $\angle 5>\angle 6$ [ Angle opposite to the longer side is greater ]

In $△CBD,$

$\Rightarrow$ $CD>BC$ [ From ( 2 ) ]

$\therefore$ $\angle 7>\angle 8$ [ Angle opposite to the longer side is greater ]

Adding equations

$\Rightarrow$ $\angle 5+\angle 7>\angle 6+\angle 8$

$\Rightarrow$ $\angle B>\angle D$ ----- Hence proved