Question
$ABC$ is a right triangle such that $AB = AC$ and bisector of angle $C$ intersects the side $AB$ at $D.$ Prove that $AC + AD = BC.$
✓
Answer
Given In quadrilateral $ABCD, AB$ is the smallest and $CD$ is largest side to find $\angle\text{B}>\angle\text{D}\text{ or }\angle\text{D}>\angle\text{B}.$
Construction Join $BD.$
Now, in $\triangle\text{ABD},\ \text{AD}>\text{AB} [$since, $AB$ is the smallest side in $ABCD]$
$\Rightarrow\ \angle{1}>\angle{3}[$ angle opposite to larger side is greater$]...(i)$
$\text{In }\triangle\text{ABCD},\ \text{CD}>\text{BC}[$ since, $CD$ is the largest side in $ABCD]$
$\Rightarrow\ \angle{2}>\angle{4}[$angle opposite to larger side is greater$]...(ii)$
On adding Eqs.$(i)$ and $(ii), $we get $\angle1+\angle2+\angle3+\angle4$
Hence, $\angle\text{B}>\angle\text{D}$
Construction Join $BD.$
Now, in $\triangle\text{ABD},\ \text{AD}>\text{AB} [$since, $AB$ is the smallest side in $ABCD]$
$\Rightarrow\ \angle{1}>\angle{3}[$ angle opposite to larger side is greater$]...(i)$
$\text{In }\triangle\text{ABCD},\ \text{CD}>\text{BC}[$ since, $CD$ is the largest side in $ABCD]$
$\Rightarrow\ \angle{2}>\angle{4}[$angle opposite to larger side is greater$]...(ii)$
On adding Eqs.$(i)$ and $(ii), $we get $\angle1+\angle2+\angle3+\angle4$
Hence, $\angle\text{B}>\angle\text{D}$
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