Question

In a triangle abc angle b is greater than C if m is the bisector of the angle BAC and a and is perpendicular to BC prove that angle m a n is equal to half Angle B minus C

Answer 1

Note. the point M will be closer to the vertex of the triangle with the smaller angle.

Let ∠CAM = ∠BAM = x

∠A + ∠B + ∠C = 180
2x + ∠B + ∠C = 180
x = 90 - ∠B/2 - ∠C/2

In the right triangle ΔCAN,
∠C + (x + ∠MAN) = 90
∠MAN = 90 - ∠C - x
= 90 - ∠C - (90 - ∠B/2 - ∠C/2)
= 90 - ∠C - 90 + ∠B/2 + ∠C/2
= ∠B/2 - ∠C/2
= (1/2) * (∠B - ∠C)