Question:
In the given figure, ∠B < ∠A and ∠C < ∠D. Show that AD > BC.
Solution:
Given: ∠B < ∠A and ∠C < ∠D
To prove: AD > BC
Proof:
In $\Delta A O B$,
$\angle B<\angle A$
$\Rightarrow A O
In $\Delta C O D$,
$\angle C<\angle D$
$\Rightarrow O D
Adding (1) and (2), we get
$A O+O D
$\therefore A D
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