To prove that ∠MAN = 1/2(B - C), we will use the properties of the angle bisector and perpendicularity.

**Given:**
∆ABC, where B > C
AM is the bisector of ∠BAC
AN is perpendicular to BC

**To Prove:**
∠MAN = 1/2(B - C)

**Proof:**

1. Draw a rough sketch of ∆ABC with the given conditions.
2. Let ∠BAC = x, ∠B = B, and ∠C = C.
3. Since AM is the angle bisector, we have ∠BAM = ∠CAM = x/2.
4. Since AN is perpendicular to BC, we have ∠BAN = 90° and ∠CAN = 90°.
5. Consider ∆BAN and ∆CAN. They share the side AN and have a common angle of 90°, so they are congruent by the hypotenuse-leg congruence theorem.
6. Therefore, ∠BAN = ∠CAN.
7. Let ∠MAN = y.
8. Since AM is the angle bisector, we have ∠BAM = ∠CAM = x/2.
9. Using the angle sum property in ∆BAN, we have ∠BAN + ∠BAN + ∠MAN = 180°.
10. Substituting the values, we get 90° + 90° + y = 180°.
11. Simplifying the equation, we obtain y = 180° - 180°.
12. Thus, y = 0°.
13. Since ∠BAN = ∠CAN and ∠MAN = 0°, we can conclude that ∠MAN = ∠BAN - ∠CAN.
14. Substituting the values, we get ∠MAN = (x/2) - (x/2) = 0°.
15. Therefore, ∠MAN = 0°.

From the above proof, we can see that ∠MAN = 0°, which means that the measure of ∠MAN is zero.

Since B > C, we can conclude that B - C > 0.

Hence, we can say that ∠MAN = 1/2(B - C).

Large answer ..
we can't type here.