The question provided in the image contains two parts based on geometry:

Problem Statement:

In the figure, $M$ and $N$ are the midpoints of $AB$ and $AC$ respectively. All produced meet at $P$. $CA'$ is a straight line.

1. (a) Prove that $\angle AP$ is the angle bisector of $\angle BAC$.
2. (b) Prove that $CA' = CP$.

Let's go through the steps to solve each part:

Part (a): Proving that $\angle AP$ is the angle bisector of $\angle BAC$.

1. By construction, $M$ and $N$ are midpoints of $AB$ and $AC$, respectively.
2. From the diagram, $P$ is the intersection of the line extending $AB$ and $AC$. We need to prove that $AP$ bisects $\angle BAC$, meaning it divides the angle equally.
3. This can be proven by showing that the two triangles $AMP$ and $ANP$ are congruent.
   • $AM = AN$ (since $M$ and $N$ are midpoints).
   • $AP$ is common to both triangles.
   • The angle subtended by the same arc in a circle is equal.
4. Using these congruent triangles, we can conclude that $\angle BAP = \angle CAP$, hence $AP$ bisects $\angle BAC$.

Part (b): Proving that $CA' = CP$.

1. We need to prove that the line segment $CA'$ (the extended line) is equal to $CP$.
2. From the properties of the figure and the construction involving circles, we can use the fact that tangents from a common external point to a circle are equal in length.
3. Therefore, $CA' = CP$ because both are tangents from the external point $C$ to the circle.

Would you like more details on any of the steps?

Here are some related questions for practice:

1. How do you prove congruent triangles in a geometry proof?
2. What is the Angle Bisector Theorem and how does it apply here?
3. How can tangents from an external point to a circle be shown as equal?
4. How do midpoints affect properties of a triangle?
5. How can you apply the properties of a cyclic quadrilateral to this problem?

Tip: In circle geometry, remember that angles subtended by the same arc are equal, which is often useful when proving angle bisectors!