geometry

Relating medians and angle bisectors:

 

In a triangle, when a median (such as BM) intersects an altitude (such as AH) inside the triangle, the median is also an angle bisector.

This is a useful property to remember for problems like this.

 

Angle relationships:

 

Since BM is a median and an angle bisector of angle B, we have:

 

∠ABM = ∠MBC (angle bisector property)

 

We are also given that angle ACB = 41 degrees.

 

Angle sum in triangle ABC:

 

The angles in any triangle add up to 180 degrees. Therefore, in triangle ABC:

 

∠ABC + ∠ACB + ∠BAC = 180°

 

Solving for ∠MBC:

 

Since BM is a median, it divides triangle ABC into two triangles with equal areas (triangle ABM and triangle MBC). Because they share the same base (BM), the ratio of their areas is equal to the ratio of their altitudes. In this case, both triangles share the same altitude (AH), so their areas must be equal.

 

Knowing that triangle ABM and triangle MBC have equal areas, we can express this using their angles:

 

1/2 * base * height (triangle ABM) = 1/2 * base * height (triangle MBC)

 

Since the base (BM) is the same for both triangles, we can simplify this to:

 

∠ABM = ∠MBC (angles are proportional to their areas in a triangle)

 

We established earlier that due to the angle bisector property, ∠ABM = ∠MBC.

 

Now, let x represent the measure of angle MBC (which is also equal to angle ABM).

 

From the angle sum property in triangle ABC:

 

∠ABC + 41° + ∠BAC = 180°

 

Since ∠ABM (or x) bisects angle B:

 

∠ABC + 41° + x = 180°

 

Substitute ∠ABM with x:

 

x + 41° + x = 180°

 

Combine like terms:

 

2x + 41° = 180°

 

Subtract 41° from both sides:

 

2x = 139°

 

Divide both sides by 2:

 

x = 69.5°