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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