$AB$ and $CD$ are respectively the smallest and longest sides of a quadrilateral $ABCD$ (see Fig). Show that $\angle A > \angle C$ and $\angle B > \angle D$.
(N/A) Let us join $AC$.
In $\Delta ABC$,since $AB$ is the smallest side of the quadrilateral $ABCD$,we have $AB < BC$ and $AB < AC$.
Specifically,$BC > AB$.
Since the angle opposite to the longer side is larger,we have $\angle BAC > \angle BCA$ ........... $(1)$
In $\Delta ACD$,since $CD$ is the longest side of the quadrilateral $ABCD$,we have $CD > AD$ and $CD > AC$.
Specifically,$CD > AD$.
Since the angle opposite to the longer side is larger,we have $\angle CAD > \angle ACD$ ........... $(2)$
Adding $(1)$ and $(2)$,we get:
$(\angle BAC + \angle CAD) > (\angle BCA + \angle ACD)$
$\Rightarrow \angle A > \angle C$
Similarly,by joining $BD$ and using the same logic for $\Delta ABD$ and $\Delta BCD$,we can show that $\angle B > \angle D$.
In $\Delta ABC$,since $AB$ is the smallest side of the quadrilateral $ABCD$,we have $AB < BC$ and $AB < AC$.
Specifically,$BC > AB$.
Since the angle opposite to the longer side is larger,we have $\angle BAC > \angle BCA$ ........... $(1)$
In $\Delta ACD$,since $CD$ is the longest side of the quadrilateral $ABCD$,we have $CD > AD$ and $CD > AC$.
Specifically,$CD > AD$.
Since the angle opposite to the longer side is larger,we have $\angle CAD > \angle ACD$ ........... $(2)$
Adding $(1)$ and $(2)$,we get:
$(\angle BAC + \angle CAD) > (\angle BCA + \angle ACD)$
$\Rightarrow \angle A > \angle C$
Similarly,by joining $BD$ and using the same logic for $\Delta ABD$ and $\Delta BCD$,we can show that $\angle B > \angle D$.
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