Given:
A circle with center O.
AB is a diameter of the circle.
C is a point on the circumference.
g is the angle formed in the semicircle.
To prove:
The angle g is always 90 degrees.
Proof:
Step 1: Draw a diagram as described in the problem statement.
Step 2: By definition, the measure of an angle in degrees is equal to the arc it intercepts divided by the radius of the circle. Let's denote the measure of angle g as ∠C.
Step 3: Since AB is a diameter, it divides the circle into two equal semicircles. Let M be the midpoint of AB.
Step 4: Consider triangle OMC. Since M is the midpoint of AB, OM is a radius of the circle.
Step 5: In triangle OMC, OM is equal in length to OC (both are radii of the same circle).Step 6: Therefore, triangle OMC is an isosceles triangle, and by definition, its base angles are equal.
Step 7: Let's denote the measure of each base angle in triangle OMC as α.
Step 8: The sum of the angles in a triangle is always 180 degrees. Therefore, in triangle OMC, we have:
∠OMC + ∠OCM + α = 180 degrees.
Step 9: Since ∠OMC and ∠OCM are equal (as they are base angles of an isosceles triangle), we can rewrite the equation as:
2∠OMC + α = 180 degrees.
Step 10: Divide both sides of the equation by 2 to isolate the measure of ∠OMC:
∠OMC + α/2 = 90 degrees.
Step 11: Notice that ∠OMC represents half of the central angle formed by the arc AC (as AC is a diameter).
Therefore, ∠OMC = ∠C.
Step 12: Substituting ∠OMC = ∠C into the equation from step 10, we get:
∠C + α/2 = 90 degrees.
Step 13: Rearrange the equation to isolate ∠C:
∠C = 90 degrees - α/2.
Step 14: The value of α/2 is a constant for a given circle and remains the same regardless of he position of point C on the circumference.
Step 15: Since the value of ∠C is always 90 degrees minus a constant (α/2), we can conclude that ∠C is always 90 degrees, irrespective of the position of point C.
Therefore, the angle in a semicircle is always 90 degrees, which completes the proof.