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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.


Answers (1)



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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