1. **Problem Statement:** Given two isosceles triangles $\triangle ABC$ and $\triangle DBC$ on the same base $BC$, with vertices $A$ and $D$ on the same side of $BC$, and $AD$ extended to intersect $BC$ at $P$, prove: (i) $\angle ABD \cong \angle ACD$ (ii) $\angle ABP \cong \angle ACP$ (iii) $AP$ bisects $\angle A$ as well as $\angle D$ (iv) $AP$ is the perpendicular bisector of $BC$ 2. **Given:** - $\triangle ABC$ and $\triangle DBC$ are isosceles with base $BC$. - $AB = AC$ and $DB = DC$ (since both are isosceles on base $BC$). - $A$ and $D$ lie on the same side of $BC$. - $AD$ extended meets $BC$ at $P$. 3. **Proof of (i): $\angle ABD \cong \angle ACD$** - Since $\triangle ABC$ is isosceles with $AB = AC$, angles opposite equal sides are equal: $$\angle ABC = \angle ACB$$ - Similarly, in $\triangle DBC$, $DB = DC$ implies: $$\angle DBC = \angle DCB$$ - Note that $\angle ABD$ and $\angle ACD$ are formed by points $A,B,D$ and $A,C,D$ respectively. - Because $A$ and $D$ are on the same side of $BC$, and $AB=AC$, $DB=DC$, the triangles $\triangle ABD$ and $\triangle ACD$ share side $AD$ and have equal sides $AB=AC$ and $DB=DC$. - By the Side-Angle-Side (SAS) criterion, $\triangle ABD \cong \triangle ACD$. - Therefore, corresponding angles $\angle ABD = \angle ACD$. 4. **Proof of (ii): $\angle ABP \cong \angle ACP$** - Point $P$ lies on $BC$ extended, so $P$ is collinear with $B$ and $C$. - Since $\triangle ABD \cong \triangle ACD$, corresponding parts are equal. - Angles $\angle ABP$ and $\angle ACP$ correspond to angles at $B$ and $C$ in these triangles. - Hence, $\angle ABP = \angle ACP$. 5. **Proof of (iii): $AP$ bisects $\angle A$ and $\angle D$** - Since $\triangle ABD \cong \triangle ACD$, $AP$ is the common side. - $AP$ lies on the line $AD$ extended to $P$ on $BC$. - Because of the congruence and symmetry, $AP$ divides $\angle A$ into two equal parts. - Similarly, $AP$ bisects $\angle D$ by the same reasoning. 6. **Proof of (iv): $AP$ is the perpendicular bisector of $BC$** - Since $AB = AC$ and $DB = DC$, points $B$ and $C$ are equidistant from $A$ and $D$. - $AP$ lies on $AD$ and intersects $BC$ at $P$. - By the symmetry and congruence of triangles, $P$ is the midpoint of $BC$. - Also, $AP$ is perpendicular to $BC$ because it bisects the base at right angles in isosceles triangles. - Therefore, $AP$ is the perpendicular bisector of $BC$. **Final answers:** (i) $\angle ABD \cong \angle ACD$ (ii) $\angle ABP \cong \angle ACP$ (iii) $AP$ bisects $\angle A$ and $\angle D$ (iv) $AP$ is the perpendicular bisector of $BC$.