Question
$ABC$ is a triangle in which $\angle\text{B}=2\angle\text{C}.$ $D$ is a point on $BC$ such that $AD$ bisects $\angle\text{BAC}$ and $\text{AB}=\text{CD}.$ Prove that $\angle\text{BAC}=72^\circ.$
✓
Answer
Given that in ABC, $\angle\text{B}=2\ \angle\text{C}$ and $D$ is a point on $BC$ such that $AD$ bisectors $\angle\text{BAC}$ and $\text{AB}=\text{CD}.$
We have to prove that $\angle\text{BAC}=72^\circ$
Now, draw the angular bisector of $\angle\text{ABC},$ which meets $AC$ in $P$. Join $PD$
Let $\text{C}=\angle\text{ACB}=\text{y}$
$\angle\text{B}=\angle\text{ABC}=2\angle\text{C}=2\text{y}$ and also
Let $\angle\text{BAD}=\angle\text{DAC}$
$\angle\text{BAC}=2\text{x}$
$[$ AD is the bisector of $\angle\text{BAC}]$
Now, in $\triangle\text{BPC},$
$\angle\text{CBP}=\text{y}$
$[$ BP is the bisector of $\angle\text{ABC}]$
$\angle\text{PCB}=\text{y}$
$\angle\text{CBP}=\angle\text{PCB}=\text{y}[\text{PC}=\text{BP]}$
Consider, $\triangle\text{ABP}$ and $\triangle\text{DCP},$
we have $\triangle\text{ABP}=\triangle\text{DCP}=\text{y}$
$\text{AB}=\text{DC}$ [Given] And $\text{PC}=\text{BP}$ [From above]
So, by $SAS$ congruence criterion,
we have $\triangle\text{ABP}\cong\triangle\text{DCP}$
Now, $\angle\text{BAP}=\angle\text{CDF}$ and $\text{AP}=\text{DP}$ [Corresponding parts of congruent triangles are equal] $\angle\text{BAP}=\angle\text{CDP}=2$ Consider, $\triangle\text{APD},$
We have $\text{AP}=\text{DP}$
$\angle\text{ADP}=\angle\text{DAP}$ But $\angle\text{DAP}=\text{x}$
$\angle\text{ADP}=\angle\text{DAP}=\text{x}$
Now, In $\triangle\text{ABD}.$
$\angle\text{ABD}+\angle\text{BAD}+\angle\text{ADB}=180^\circ$ And also $\angle\text{ADB}+\angle\text{ADC}=180^\circ$ [Straight angle] From the above two equations,
we get $\angle\text{ABD}+\angle\text{BAD}+\angle\text{ADB}+\angle\text{ADC}$
$2\text{y}+\text{x}=\angle\text{ADP}+\angle\text{PDC}$
$2\text{y}+\text{x}=\text{x}+2\text{y}$
$2\text{y}=2\text{x}$
$\text{y}=\text{x}\text{ (or})\text{ x}=\text{y}$ We know, Sum of angles in a triangle $= 180^\circ$
So, In $\triangle\text{ABC},$
$\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ$
$2\text{x}+2\text{y}+\text{y}=180^\circ$
$[\angle\text{A}=2\text{x},\angle\text{B}=2\text{y},\angle\text{C}=\text{y}]$
$2(\text{y}+3\text{y}=180^\circ[\text{x}=\text{y}]$
$5\text{y}=180^\circ$
$\text{y}=36^\circ$ Now, $\angle\text{A}=\angle\text{BAC}=2\times36^\circ=72^\circ$
We have to prove that $\angle\text{BAC}=72^\circ$
Now, draw the angular bisector of $\angle\text{ABC},$ which meets $AC$ in $P$. Join $PD$
Let $\text{C}=\angle\text{ACB}=\text{y}$
$\angle\text{B}=\angle\text{ABC}=2\angle\text{C}=2\text{y}$ and also
Let $\angle\text{BAD}=\angle\text{DAC}$
$\angle\text{BAC}=2\text{x}$
$[$ AD is the bisector of $\angle\text{BAC}]$
Now, in $\triangle\text{BPC},$
$\angle\text{CBP}=\text{y}$
$[$ BP is the bisector of $\angle\text{ABC}]$
$\angle\text{PCB}=\text{y}$
$\angle\text{CBP}=\angle\text{PCB}=\text{y}[\text{PC}=\text{BP]}$
Consider, $\triangle\text{ABP}$ and $\triangle\text{DCP},$
we have $\triangle\text{ABP}=\triangle\text{DCP}=\text{y}$
$\text{AB}=\text{DC}$ [Given] And $\text{PC}=\text{BP}$ [From above]
So, by $SAS$ congruence criterion,
we have $\triangle\text{ABP}\cong\triangle\text{DCP}$
Now, $\angle\text{BAP}=\angle\text{CDF}$ and $\text{AP}=\text{DP}$ [Corresponding parts of congruent triangles are equal] $\angle\text{BAP}=\angle\text{CDP}=2$ Consider, $\triangle\text{APD},$
We have $\text{AP}=\text{DP}$
$\angle\text{ADP}=\angle\text{DAP}$ But $\angle\text{DAP}=\text{x}$
$\angle\text{ADP}=\angle\text{DAP}=\text{x}$
Now, In $\triangle\text{ABD}.$
$\angle\text{ABD}+\angle\text{BAD}+\angle\text{ADB}=180^\circ$ And also $\angle\text{ADB}+\angle\text{ADC}=180^\circ$ [Straight angle] From the above two equations,
we get $\angle\text{ABD}+\angle\text{BAD}+\angle\text{ADB}+\angle\text{ADC}$
$2\text{y}+\text{x}=\angle\text{ADP}+\angle\text{PDC}$
$2\text{y}+\text{x}=\text{x}+2\text{y}$
$2\text{y}=2\text{x}$
$\text{y}=\text{x}\text{ (or})\text{ x}=\text{y}$ We know, Sum of angles in a triangle $= 180^\circ$
So, In $\triangle\text{ABC},$
$\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ$
$2\text{x}+2\text{y}+\text{y}=180^\circ$
$[\angle\text{A}=2\text{x},\angle\text{B}=2\text{y},\angle\text{C}=\text{y}]$
$2(\text{y}+3\text{y}=180^\circ[\text{x}=\text{y}]$
$5\text{y}=180^\circ$
$\text{y}=36^\circ$ Now, $\angle\text{A}=\angle\text{BAC}=2\times36^\circ=72^\circ$
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