In the shown figure,  O is the centre of the circle.  PB  =  PC, ∠PBO  =  25∘ and  ∠BOC  =  130∘, then find  ∠ABP + ∠DCP.

1. Break Down the Problem

To find $\angle ABP + \angle DCP$, we start by identifying the given variables and constructing relationships between the angles.

2. Relevant Concepts

1. Since $O$ is the center of the circle, $PB = PC$ means that triangle $PBC$ is isosceles with $PB = PC$.
2. The angles $\angle PBO$ and $\angle PCO$ are equal because they subtend the same arc $BC$ in the circle.
3. The relationship between the angles can be established using properties of triangles and the fact that the sum of angles in a triangle is $180^\circ$.

3. Analysis and Detail

1. Finding $\angle POC$: $\angle POC = 2 \times \angle PBO = 2 \times 25^\circ = 50^\circ$

2. Finding $\angle BOC$: Since $\angle BOC = 130^\circ$, we now know the angle $\angle POC$ in triangle $POB$ and $POC$: $\angle BOP = 130^\circ - 50^\circ = 80^\circ$

3. Using Triangle Properties: Since $PBC$ is isosceles: $\angle PBC + \angle PCB + \angle BPC = 180^\circ$ Given that: $\angle PBC = \angle PCB \quad (\text{Let } \angle PBC = x)$ Then: $x + x + 50^\circ = 180^\circ \implies 2x = 130^\circ \implies x = 65^\circ$ Therefore, $\angle PBC = \angle PCB = 65^\circ$.

4. Finding $\angle ABP + \angle DCP$: Since $ABP$ and $DCP$ are vertical angles to the angles we have: $\angle ABP = \angle PBC = 65^\circ$ $\angle DCP = \angle PCB = 65^\circ$ Combining these: $\angle ABP + \angle DCP = 65^\circ + 65^\circ = 130^\circ$

4. Verify and Summarize

Each angle has been calculated through isosceles triangle properties and the relationships between the angles based on the triangle summation property.

Final Answer

$\angle ABP + \angle DCP = 130^\circ$