Question

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

Answer 1

Let $ABCD$ be a quadrilateral such that $AB$ is it's smallest side and $CD$ is it largest side.

Now join $AC$ and $BD$

In $\Delta ABC$ ,we have

$BC>AB$

$\Rightarrow \angle 8>\angle 3$ ...(1) {$\because$ angle opposite to longer side is greater.}

Since $CD$ is the longest side of quadrilateral $ABCD$

In $\Delta ACD$, we have

$CD>AD$

$\Rightarrow \angle 7>\angle 4$ ...(2)

Adding (1) and(2) we get

$\angle 8+\angle 7>\angle 3+\angle 4$

$\Rightarrow \angle A>\angle C$

Now in $\Delta ABD$, we have

$AD>AB$ {$\because AB$ is the shortest side}

$\Rightarrow \angle 1>\angle 6$ ...(3)

In $\Delta BCD$,we have

$CD>BC$

$\Rightarrow \angle 2>\angle 5$ ....(4)

Adding (3) and (4)

$\angle 1+\angle 2>\angle 5+\angle 6$

$\Rightarrow \angle B>\angle D$

Hence proved.