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 right angled $\triangle\text{ABC},\text{AB}= \text{AC}\text{ and }\text{CD}$ is the bisector of $\angle\text{C}.$ contruction draw $\text{DE}\ \bot\ \text{BC}.$ to prove $\text{AC}+\text{AD}= \text{BC}$ proof in right angled $\triangle\text{ABC},\text{AB}=\text{AC}\text{ and }\text{BC}$ is a hypotenuse [given]
$\angle1=\angle2$ [given,$\text{CD}$ isthe bisector of $\angle\text{C}$] $\text{DC}=\text{DC}$ [common sides] $\therefore\ \triangle\text{DAC}\cong\triangle\text{DEC}$ [by AAS congruence rule] $\Rightarrow\ \text{DA}=\text{DE}.$ [by CPCT]...(i) $\text{and}\ \text{AC}=\text{EC}$ ...(ii) $\text{ln}\ \triangle\text{ABC},\ \text{AB}=\text{AC}$ $\angle\text{C}=\angle{B}$ [angle opposite to equal sides are equal]...(iii) Again,in $\triangle\text{ABC},\ \angle\text{A}+\angle\text{B}+\angle\text{C}= 180^\circ$ [by angle sum property of a triangle] $\Rightarrow\ 90^\circ+\angle\text{B}+\angle\text{B}=180^\circ$ $\Rightarrow\ 2\angle\text{B}=180^\circ-90^\circ$[from Eq.(iii)] $\Rightarrow\ 2\angle\text{B}=90^\circ$ $\Rightarrow\ \angle\text{B}=45^\circ$ $\text{ln}\ \triangle\text{BED},\ \angle5=180^\circ-\big(\angle\text{B}+\angle{4}\big)$ [by angle sum property of a triangle] $=180^\circ-(45^\circ+90^\circ)$ $=180^\circ-135^\circ=45^\circ$ $\therefore\ \angle\text{B}=\angle{5}$ $\Rightarrow\ \text{DE}=\text{BE}\ [\because$ sides opposite to equal angle are equal$]$ ...(iv) from Eqs. (i) and (iv), $\text{DA}=\text{DE}=\text{BE}$ ...(v) $\because\text{BC}=\text{CE}+\text{EB}$ $=\text{CA}+\text{DA}$ [from Eqs. (ii) and (v)] $\therefore\text{AD}+\text{AC}=\text{BC}$ Hence proved.
$\angle1=\angle2$ [given,$\text{CD}$ isthe bisector of $\angle\text{C}$] $\text{DC}=\text{DC}$ [common sides] $\therefore\ \triangle\text{DAC}\cong\triangle\text{DEC}$ [by AAS congruence rule] $\Rightarrow\ \text{DA}=\text{DE}.$ [by CPCT]...(i) $\text{and}\ \text{AC}=\text{EC}$ ...(ii) $\text{ln}\ \triangle\text{ABC},\ \text{AB}=\text{AC}$ $\angle\text{C}=\angle{B}$ [angle opposite to equal sides are equal]...(iii) Again,in $\triangle\text{ABC},\ \angle\text{A}+\angle\text{B}+\angle\text{C}= 180^\circ$ [by angle sum property of a triangle] $\Rightarrow\ 90^\circ+\angle\text{B}+\angle\text{B}=180^\circ$ $\Rightarrow\ 2\angle\text{B}=180^\circ-90^\circ$[from Eq.(iii)] $\Rightarrow\ 2\angle\text{B}=90^\circ$ $\Rightarrow\ \angle\text{B}=45^\circ$ $\text{ln}\ \triangle\text{BED},\ \angle5=180^\circ-\big(\angle\text{B}+\angle{4}\big)$ [by angle sum property of a triangle] $=180^\circ-(45^\circ+90^\circ)$ $=180^\circ-135^\circ=45^\circ$ $\therefore\ \angle\text{B}=\angle{5}$ $\Rightarrow\ \text{DE}=\text{BE}\ [\because$ sides opposite to equal angle are equal$]$ ...(iv) from Eqs. (i) and (iv), $\text{DA}=\text{DE}=\text{BE}$ ...(v) $\because\text{BC}=\text{CE}+\text{EB}$ $=\text{CA}+\text{DA}$ [from Eqs. (ii) and (v)] $\therefore\text{AD}+\text{AC}=\text{BC}$ Hence proved.
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