AB and CD are respectively the smallest and largest sides of a quadrilateral ABCD. Show that ∠A > ∠C and
∠B > ∠D.
Given: In quadrilateral ABCD, AB and CD are respectively the smallest and largest sides.
To prove:
(i) ∠A > ∠C
(ii) ∠B > ∠D
Construction: Join AC.
Proof:
In $\Delta A B C$,
$\because B C>A B \quad$ (Given, $A B$ is the smallest side)
$\therefore \angle 1>\angle 2 \quad \ldots(1)$
In $\Delta A D C$,
$\because C D>A D \quad$ (Given, $C D$ is the largest side)
$\therefore \angle 3>\angle 4 \quad \ldots(2)$
Adding (1) and (2), we get
$\angle 1+\angle 3>\angle 2+\angle 4$
$\therefore \angle A>\angle C$
(ii)
Construction: Join BD.
Proof:
In $\Delta A B D$,
$\because A D>A B \quad$ (Given, $A B$ is the smallest side.)
$\therefore \angle 5>\angle 6 \quad \ldots(3)$
In $\Delta C B D$,
$\because C D>B C \quad$ (Given, $C D$ is the greatest side.)
$\therefore \angle 7>\angle 8 \quad \ldots(4)$
Adding $(3)$ and $(4)$, we get
$\angle 5+\angle 7>\angle 6+\angle 8$
$\therefore \angle B>\angle D$
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