$ABC$ is an isosceles triangle with $AB = AC$. Draw $AP \perp BC$ to show that $\angle B = \angle C$.
(N/A) Given: $ABC$ is an isosceles triangle with $AB = AC$. $AP \perp BC$.
To prove: $\angle B = \angle C$.
Proof:
In $\Delta ABP$ and $\Delta ACP$:
$1$. $\angle APB = \angle APC = 90^\circ$ (Since $AP \perp BC$)
$2$. $AB = AC$ (Given)
$3$. $AP = AP$ (Common side)
Therefore,by $RHS$ congruence criterion,$\Delta ABP \cong \Delta ACP$.
Since the triangles are congruent,their corresponding parts are equal $(CPCT)$.
Thus,$\angle B = \angle C$.
To prove: $\angle B = \angle C$.
Proof:
In $\Delta ABP$ and $\Delta ACP$:
$1$. $\angle APB = \angle APC = 90^\circ$ (Since $AP \perp BC$)
$2$. $AB = AC$ (Given)
$3$. $AP = AP$ (Common side)
Therefore,by $RHS$ congruence criterion,$\Delta ABP \cong \Delta ACP$.
Since the triangles are congruent,their corresponding parts are equal $(CPCT)$.
Thus,$\angle B = \angle C$.
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Similar Questions
$\Delta ABC$ and $\Delta DBC$ are two isosceles triangles on the same base $BC$ and vertices $A$ and $D$ are on the same side of $BC$ (see Fig). If $AD$ is extended to intersect $BC$ at $P$,show that:
$(i)$ $\Delta ABD \cong \Delta ACD$
$(ii)$ $\Delta ABP \cong \Delta ACP$
$(iii)$ $AP$ bisects $\angle A$ as well as $\angle D$.
$(iv)$ $AP$ is the perpendicular bisector of $BC$.
$(i)$ $\Delta ABD \cong \Delta ACD$
$(ii)$ $\Delta ABP \cong \Delta ACP$
$(iii)$ $AP$ bisects $\angle A$ as well as $\angle D$.
$(iv)$ $AP$ is the perpendicular bisector of $BC$.
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$(i)$ $\Delta AMC \cong \Delta BMD$
$(ii)$ $\angle DBC$ is a right angle.
$(iii)$ $\Delta DBC \cong \Delta ACB$
$(iv)$ $CM = \frac{1}{2} AB$
$(i)$ $\Delta AMC \cong \Delta BMD$
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$(iii)$ $\Delta DBC \cong \Delta ACB$
$(iv)$ $CM = \frac{1}{2} AB$
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