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$ \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} $.
"
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,
$\angle A > \angle C$.
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