The problem describes a triangle ABC where sides AC and BC are equal. Ray CM is the bisector of the external angle BCD, and angle MCD is 57 degrees. Find the measure of angle BAC. The answer should be in degrees.

Solution:

• Given that AC = BC, triangle ABC is an isosceles triangle.
• CM is the bisector of the external angle BCD.
• Angle MCD = 57°.
• Since CM bisects angle BCD, angle BCM = angle MCD = 57°.
• Therefore, angle BCD = angle BCM + angle MCD = 57° + 57° = 114°.
• The angle BCD and the internal angle ACB form a linear pair, so their sum is 180°.
• Angle ACB = 180° - angle BCD = 180° - 114° = 66°.
• In isosceles triangle ABC, AC = BC, so the angles opposite these sides are equal: angle BAC = angle ABC.
• The sum of angles in triangle ABC is 180°.
• Angle BAC + angle ABC + angle ACB = 180°.
• Since angle BAC = angle ABC, we can write: 2 * angle BAC + angle ACB = 180°.
• 2 * angle BAC + 66° = 180°.
• 2 * angle BAC = 180° - 66° = 114°.
• Angle BAC = 114° / 2 = 57°.

Answer: 57