$ABC$ 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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$\Delta ABC$ 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:
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