ABC is a right triangle such that AB = AC and bisector of angle C…

Question

$ABC$ is a right triangle such that $AB = AC$ and bisector of angle $C$ intersects the side $AB$ at $D.$ Prove that $AC + AD = BC.$

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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.