Question

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

Answer 1

$ABCD$ is a quadrilateral such that $AB$ is its smallest side and $CD$ is its largest side.

Join $AC$ and $BD.$ Since $AB$ is the smallest side of quadrilateral $ABCD.$

$\therefore$ In $△ABC$, we have

$BC>AB$

$\Rightarrow \angle 8>\angle 3$ ... (1)

[Since angle opposite to longer side is greater]

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

Now, In $△ACD$, we have

$CD>AD$

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

[Since angle opposite to longer side is greater]

Adding (1) and (2), we get

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

$\Rightarrow \angle A>\angle C$

Again, in $△ABD$, we have

$AD>AB$ [Since $AB$ is the shortest side]

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

In $△BCD$, we have

$CD>BC$ [Since $CD$ is the longest side]

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

Adding (3) and (4), we get

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

$\Rightarrow \angle B>\angle D$

Thus , $\angle A>\angle C$ and $\angle B>\angle D.$ $\left[henceproved\right]$