In $\Delta ABC$,the bisector $AD$ of $\angle A$ is perpendicular to side $BC$ (see figure). Show that $AB = AC$ and $\Delta ABC$ is isosceles.
(N/A) In $\Delta ABD$ and $\Delta ACD$:
$\angle BAD = \angle CAD$ (Given,as $AD$ is the bisector of $\angle A$)
$AD = AD$ (Common side)
$\angle ADB = \angle ADC = 90^\circ$ (Given,as $AD \perp BC$)
Therefore,by the $ASA$ (Angle-Side-Angle) congruence rule,$\Delta ABD \cong \Delta ACD$.
Since the triangles are congruent,their corresponding parts are equal $(CPCT)$.
Thus,$AB = AC$.
Since two sides of $\Delta ABC$ are equal,$\Delta ABC$ is an isosceles triangle.
$\angle BAD = \angle CAD$ (Given,as $AD$ is the bisector of $\angle A$)
$AD = AD$ (Common side)
$\angle ADB = \angle ADC = 90^\circ$ (Given,as $AD \perp BC$)
Therefore,by the $ASA$ (Angle-Side-Angle) congruence rule,$\Delta ABD \cong \Delta ACD$.
Since the triangles are congruent,their corresponding parts are equal $(CPCT)$.
Thus,$AB = AC$.
Since two sides of $\Delta ABC$ are equal,$\Delta ABC$ is an isosceles triangle.
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