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

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Consider $\Delta ADC$ in the above figure :

$AD\ <\ CD$ (Given)

Thus $\angle CAD\ > \angle ACD$ (as angle opposite to smaller side is smaller)

Now consider $\Delta ABC$ ,

We have : $BC\ > AB$

and $\angle BAC\ > \angle ACB$

Adding the above result we get,

$\angle BAC\ +\ \angle CAD > \angle ACB\ +\ \angle ACD$

or $\small \angle A>\angle C$

Similarly, consider $\Delta ABD$ ,

we have $AB\ <\ AD$

Therefore $\angle ABD\ > \angle ADB$

and in $\Delta BDC$ we have,

$CD\ >\ BC$

and $\angle CBD\ >\ \angle CDB$

from the above result we have,

$\angle ABD\ +\ \angle CBD\ >\ \angle ADB\ +\ \angle CDB$

or $\small \angle B>\angle D$

Hence proved.

Posted by

Sanket Gandhi

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