Question
$ABC$ and $DBC$ are two triangles on the same base $BC$ such that $A and D$ lie on the opposite sides of $BC, AB = AC$ and $DB = DC.$ Show that $AD$ is the perpendicular bisector of $BC.$
✓
Answer
Given two $\triangle\text{ABC}$ and $\angle\text{DBC}$ are formed on the same base $BC$ Such that $A$ and $D$ lie on the opposite side of $AB = AC$ and $DB = DC.$
In $\triangle\text{ABD}$ and $\triangle\text{ACD},$
$\text{AB}=\text{AC}$
$\angle\text{ABD}=\angle\text{ACD}$
$\angle\text{BAD}=\angle\text{CAD}$
In $\triangle\text{ABO},$
$\text{AB}=\text{AC}$
$\angle\text{AOB}+\angle\text{AOB}=180^{\circ}$
$2\angle\text{AOB}=180^{\circ}$
$\angle\text{AOB}=\frac{180^{\circ}}{2}=180^{\circ}$
Hence, $AD$ is the perpendicular of $BC.$
In $\triangle\text{ABD}$ and $\triangle\text{ACD},$
$\text{AB}=\text{AC}$
$\angle\text{ABD}=\angle\text{ACD}$
$\angle\text{BAD}=\angle\text{CAD}$
In $\triangle\text{ABO},$
$\text{AB}=\text{AC}$
$\angle\text{AOB}+\angle\text{AOB}=180^{\circ}$
$2\angle\text{AOB}=180^{\circ}$
$\angle\text{AOB}=\frac{180^{\circ}}{2}=180^{\circ}$
Hence, $AD$ is the perpendicular of $BC.$
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