# $\mathrm{AB}$ 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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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\$.

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