Go To Your Study Dashboard

Get Answers to all your Questions

header-bg

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 $\small \angle B>\angle D$ .

Answers (1)

best_answer

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,

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

$\small \angle A>\angle C$

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,

$\small \angle B>\angle D$

Hence proved.

Posted by

Sanket Gandhi

View full answer

Similar Questions

Trending Articles/News

anam-khan

Board Exam Date 2025 Live: Bihar Class 10, 12 model papers out, time table soon; updates on CBSE, UP

Latest Question