Question

Question 4
AB and CD are respectively the smallest and longest sides of a quadrilateral ABCD (see the given figure). Show that $\angle A$ > $\angle C$ and $\angle B$ > $\angle D$.

Answer 1

Let us join AC.
$In\Delta ABC$,
AB < BC (AB is the smallest side of quadrilateral ABCD)
$\therefore \angle 2<\angle 1\dots \left(i\right)$ (Angle opposite to the smaller side is smaller)
In $\Delta$ADC,
AD < CD (CD is the largest side of quadrilateral ABCD)
$\therefore \angle 4<\angle 3\dots \left(ii\right)$ (Angle opposite to the smaller side is smaller)
On adding equations (i) and (ii), we obtain
$\angle 2+\angle 4<\angle 1+\angle 3$
$\Rightarrow \angle C<\angle A$
$\Rightarrow \angle A>\angle C$
Let us join BD

$In\Delta ABD$,
AB < AD (AB is the smallest side of quadrilateral ABCD)
$\therefore \angle 8<\angle 5\dots \left(iii\right)$ (Angle opposite to the smaller side is smaller)
$In\Delta BDC,$
BC < CD (CD is the largest side of quadrilateral ABCD)
$\therefore \angle 7<\angle 6\dots \left(iv\right)$ (Angle opposite to the smaller side is smaller)
On adding equations (iii) and (iv), we obtain
$\angle 8+\angle 7<\angle 5+\angle 6$
$\Rightarrow \angle D<\angle B$
$\Rightarrow \angle B>\angle D$