M is a point on side AB of a triangle ABC such that AM = BM = CM. If angles A…

If a point on one side of a triangle is equidistant from all three vertices, that point is the circumcentre of the triangle and that side is a diameter of its circumcircle. By Thales' theorem — the angle in a semicircle is a right angle — the angle subtended by a diameter at any point on the circle is always a right angle.

1. Since AM = BM, M is the midpoint of side AB; since AM = BM = CM as well, M is equidistant from A, B, and C, so M is the circumcentre of triangle ABC and AB is a diameter of its circumcircle.

2. By Thales' theorem, the angle at C subtended by the diameter AB is a right angle, so angle ACB = 90°.

3. Apply the angle-sum property of a triangle: angle A + angle B + angle C = 180°, that is, x + 70° + 90° = 180°.

4. Solve for x: x = 180° − 160° = 20°.

5. Substitute x into the expression: 3x + 25° = 3(20°) + 25° = 60° + 25° = 85°.

The same value can be confirmed independently, without invoking Thales' theorem, using only the isosceles triangles formed at M.

1. Since AM = CM, triangle AMC is isosceles, so angle ACM = angle CAM = x.

2. Since BM = CM, triangle BMC is isosceles, so angle BCM = angle CBM = 70°.

3. Therefore angle C = angle ACM + angle BCM = x + 70°.

4. Substitute this into the angle-sum property: x + 70° + (x + 70°) = 180°, that is 2x + 140° = 180°.

5. Solve for x: 2x = 40°, so x = 20° — the same value found above, which confirms that 3x + 25° = 85°.