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

Given: ABCD is a quadrilateral having AB as the smallest side and CD as the longest side

To prove: \(\angle{A}\) > \(\angle{C}\) and \(\angle{B}\) > \(\angle{D}\)

Construction: Join A to C and B to D.

Figure:

Proof:
In \(\triangle{ABC}\), we have AB is the smallest side.

As, AB < BC
\(\therefore \) \(\angle{5}\) < \(\angle{1}\) ...(i)
(Since, angle opposite to longer side is greater)

Similarly, in \(\triangle{ADC}\),
we have CD is the largest side.

As, AD < CD
\(\therefore \) \(\angle{6}\) < \(\angle{2}\) ...(ii)
(Since, angle opposite to longer side is greater)

On adding Equation (i) and (ii), we get,

\(\Rightarrow \) \(\angle{5}\) + \(\angle{6}\) < \(\angle{1}\) + \(\angle{2}\)
\(\Rightarrow \) \(\angle{C}\) < \(\angle{A}\)
\(\Rightarrow \) \(\angle{A}\) > \(\angle{C}\)

Now, Similarly, \(\triangle{ADB}\), we have AB is the smallest side.
As, AD > AB
\(\therefore \) \(\angle{3}\) > \(\angle{8}\) ...(iii)
(Since, angle opposite to longer side is greater)

Similarly, in \(\triangle{BCD}\), we have CD is the largest side.
As, CD > BC
\(\therefore \) \(\angle{4}\) > \(\angle{7}\) ...(iv)
(Since, angle opposite to longer side is greater)

On adding Equation (iii) and (iv), we get,
\(\Rightarrow \) \(\angle{3}\) + \(\angle{4}\) > \(\angle{8}\) + \(\angle{7}\)
\(\Rightarrow \) \(\angle{B}\) > \(\angle{D}\)
Hence, proved.