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# 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.