In $\Delta ABC$,$AD$ is the perpendicular bisector of $BC$ (see figure). Show that $\Delta ABC$ is an isosceles triangle in which $AB = AC$.
(N/A) Given: $AD$ is the perpendicular bisector of $BC$,which implies $BD = CD$ and $\angle ADB = \angle ADC = 90^o$.
In $\Delta ABD$ and $\Delta ACD$:
$AD = AD$ (Common side)
$\angle ADB = \angle ADC = 90^o$ (Given)
$BD = CD$ (Since $AD$ is the bisector of $BC$)
Therefore,$\Delta ABD \cong \Delta ACD$ by the $SAS$ congruence criterion.
Since the triangles are congruent,their corresponding parts are equal $(CPCT)$.
Thus,$AB = AC$.
Since two sides of $\Delta ABC$ are equal,it is an isosceles triangle.
In $\Delta ABD$ and $\Delta ACD$:
$AD = AD$ (Common side)
$\angle ADB = \angle ADC = 90^o$ (Given)
$BD = CD$ (Since $AD$ is the bisector of $BC$)
Therefore,$\Delta ABD \cong \Delta ACD$ by the $SAS$ congruence criterion.
Since the triangles are congruent,their corresponding parts are equal $(CPCT)$.
Thus,$AB = AC$.
Since two sides of $\Delta ABC$ are equal,it is an isosceles triangle.
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