The circle and the lines AC & AB are illustrations of congruent angles
See below for the proof that the angles ∠ACB and ∠MAB are equal
How to prove that ∠MAB=∠ACB
To prove that ∠MAB=∠ACB, we make use of the attached diagram
From the attached diagram, we have:
∠MAB = x°
The angle at between a tangent and a radius or a diameter is a right angle.
This means that:
∠MAB + ∠BAC = 90°
Substitute ∠MAB = x°
x° + ∠BAC = 90°
Subtract x from both sides
∠BAC = 90° - x°
The angle at in a semicircle is a right angle.
This means that:
∠CBA = 90°
The angles in a triangle add up to 180°.
So, we have:
∠BAC + ∠CBA + ∠ACB = 180°
Substitute ∠CBA = 90° and ∠BAC = 90° - x°
90° - x° + 90° + ∠ACB = 180°
Evaluate the like terms
180° - x° + ∠ACB = 180°
Subtract 180° from both sides
- x° + ∠ACB = 0
Add x° to both sides
∠ACB = x°
Recall that:
∠MAB = x°
So, we have:
∠ACB = ∠MAB = x°
Hence, the angles ∠ACB and ∠MAB are equal
Read more about congruent angles at:
https://brainly.com/question/1675117
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