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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!

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

(a) Prove that AP is the angle bisector of ∠BAC.
(b) Prove that CA' = CP.

Learn how to prove that AP is the angle bisector of ∠BAC and that CA' = CP using concepts from circle geometry, congruent triangles, and the Angle Bisector Theorem. Suitable for high school students in Grades 9-11.

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