Question
In the given figure, $\angle\text{B}<\angle\text{A}$ and $\angle\text{C}<\angle\text{D}.$ Show that $\text{AD < BC}.$
✓
Answer
Given: $\angle\text{B}<\angle\text{A}$ and $\angle\text{C}<\angle\text{D}$
To prove: $\text{AD > BC}$
Proof: In $\triangle\text{AOB},$
$\angle\text{B}<\angle\text{A}$
$\Rightarrow\text{AO}<\text{BO}$ $($Side opposite to the greater angle is longer$)...(1)$
In $\triangle\text{COD},$ $\angle\text{C}<\angle\text{D}$
$\Rightarrow\text{OD < OC}$ (Side opposite to the greater angle is longer$)...(2)$ Adding $(1)$ and $(2),$
we get $\text{AO + OD < BO + OC}$
$\therefore\text{AD < BC}$
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