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Answer :

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.

- Show that in a right angled triangle, the hypotenuse is the longest side.
- In figure, sides AB and AC of are extended to points P and Q respectively. Also, \(\angle{PBC}\) AB.
- In figure, \(\angle{B}\)
- In figure, PR > PQ and PS bisects \(\angle{QPR}\). Prove that \(\angle{PSR}\) > \(\angle{PSQ}\).
- Show that of all line segments drawn from a give point not on it, the perpendicular line segment is the shortest.

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