$AD$ is an altitude of an isosceles triangle $ABC$ in which $AB = AC$. Show that:
$(i)$ $AD$ bisects $BC$
$(ii)$ $AD$ bisects $\angle A$.
$(i)$ $AD$ bisects $BC$
$(ii)$ $AD$ bisects $\angle A$.
(N/A) $(i)$ In $\Delta ABD$ and $\Delta ACD$,we have:
$AB = AC$ [Given]
$\angle ADB = \angle ADC = 90^\circ$ [Since $AD$ is an altitude]
$AD = AD$ [Common side]
Therefore,by $RHS$ congruence rule,$\Delta ABD \cong \Delta ACD$.
Since the triangles are congruent,their corresponding parts are equal $(CPCT)$.
Therefore,$BD = CD$,which means $AD$ bisects $BC$.
$(ii)$ Since $\Delta ABD \cong \Delta ACD$,their corresponding angles are equal $(CPCT)$.
Therefore,$\angle BAD = \angle CAD$,which means $AD$ bisects $\angle A$.
$AB = AC$ [Given]
$\angle ADB = \angle ADC = 90^\circ$ [Since $AD$ is an altitude]
$AD = AD$ [Common side]
Therefore,by $RHS$ congruence rule,$\Delta ABD \cong \Delta ACD$.
Since the triangles are congruent,their corresponding parts are equal $(CPCT)$.
Therefore,$BD = CD$,which means $AD$ bisects $BC$.
$(ii)$ Since $\Delta ABD \cong \Delta ACD$,their corresponding angles are equal $(CPCT)$.
Therefore,$\angle BAD = \angle CAD$,which means $AD$ bisects $\angle A$.
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