Question
In the given figure, $\text{AD}\perp\text{BC}$ and CD > BD. Show that AC > AB.
✓
Answer
Given: $\text{AD}\perp\text{BC}$ and $\text{CD > BD}$ To prove: $\text{AC > AB}$ Proof:$\angle\text{ADB}=\angle\text{ADC}=90^{\circ}$ $(\text{AD}\perp\text{BC})...(1)$$\angle\text{BAD}<\angle\text{DAC}$ $(\text{CD}>\text{BD})...(2)$
In $\triangle\text{}\text{ABD},$ Using angle sum property of a triangle$\angle\text{B}=180^{\circ}-\angle\text{ADB}-\angle\text{BAD}$
$\angle\text{B}=90^{\circ}-\angle\text{BAD}...(3)$
In $\triangle\text{ADC,}$ Using angle sum property of a triangle,$\angle\text{ACD}=90^{\circ}-\angle\text{DAC}...(4)$
From (2), (3) and (4), we get$\angle\text{B}>\angle\text{C}$
Therefore, $\text{AC}>\text{AB}.$
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