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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