$ \mathrm{AB} $ and $ \mathrm{CD} $ are respectively the smallest and longest sides of a quadrilateral $ \mathrm{ABCD} $ (see Fig. 7.50). Show that $ \angle A>\angle C $ and $ \angle \mathrm{B}>\angle \mathrm{D} $.
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AcademicMathematicsNCERTClass 9

Given:

$AB$ and $CD$ are respectively the smallest and longest sides of a quadrilateral $ABCD$.

To do:

We have to show that $\angle A>\angle C$ and $\angle B>\angle $D$.

Solution:


Let us consider $\triangle ABD$, We have,

$AB

We know that,

The angle opposite the longer side will always be larger.

This implies,

$\angle ADB

In a similar way in $\triangle BCD$,

We have,

$BC

This implies,

$\angle BDC

By adding (i) and (ii) we get,

$\angle ADB +\angle BDC

This implies,

$\angle ADC

$\angle B > \angle D$

Similarly, In triangle $ABC$,

We know that the angle opposite to the longer side will always be larger.

$\angle ACB

In a similar way from $\triangle ADC$,

We get,

$\angle DCA

By adding (iii) and (iv) we get,

$\angle ACB + \angle DCA

This implies,

$\angle BCD

Therefore,

$\angle A > \angle C$.

raja
Updated on 10-Oct-2022 13:41:23

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