Young Math Brains

Angle in a semicircle: complete theorem proof and exam method

6 min read

Question

AB is a diameter of a circle. Point C lies anywhere on the semicircle. Find ∠ACB and prove that its value does not change as C moves.

Angle in a semicircle: The mistake that costs marksWatch on YouTube

We need to determine ∠ACB\angle ACB without measuring the drawing. The proof below starts from the equal radii, expresses all three angles of triangle ABC in terms of x and y, and only then evaluates the required angle.

AOBCxyAB is a diameter
Joining C to the centre creates two isosceles triangles. Their base angles split the angle at C into x and y. The triangle-angle sum below will determine x + y before the required angle is stated.

Step 1: create two isosceles triangles

Let OO be the centre of the circle. Join OO to point CC. Since OAOA, OBOB and OCOC are radii of the same circle, they are equal in length (OA=OB=OCOA = OB = OC).

This splits △ABC\triangle ABC into two isosceles triangles: △AOC\triangle AOC and △BOC\triangle BOC. Let ∠ACO=x\angle ACO = x and ∠OCB=y\angle OCB = y:

In △AOC:OA=OC(Radii of the same circle)\text{In } \triangle AOC: OA = OC \quad (\text{Radii of the same circle})
  ⟹  ∠OAC=∠ACO=x(Angles opposite to equal sides are equal)\implies \angle OAC = \angle ACO = x \quad (\text{Angles opposite to equal sides are equal})
In △BOC:OB=OC(Radii of the same circle)\text{In } \triangle BOC: OB = OC \quad (\text{Radii of the same circle})
  ⟹  ∠OBC=∠OCB=y(Angles opposite to equal sides are equal)\implies \angle OBC = \angle OCB = y \quad (\text{Angles opposite to equal sides are equal})
Since A, O, B lie on diameter AB:∠BAC=x and ∠ABC=y\text{Since } A\text{, } O\text{, } B \text{ lie on diameter } AB: \angle BAC = x \text{ and } \angle ABC = y
At vertex C:∠ACB=∠ACO+∠OCB\text{At vertex } C: \angle ACB = \angle ACO + \angle OCB
∠ACB=x+y\angle ACB = x + y

Step 2: use the angle sum of triangle ABC

Now apply the Angle Sum Property to the entire triangle △ABC\triangle ABC, substituting the angle expressions derived in Step 1:

∠BAC+∠ACB+∠ABC=180∘(Angle sum property of △ABC)\angle BAC + \angle ACB + \angle ABC = 180^\circ \quad (\text{Angle sum property of } \triangle ABC)
Substitute angle values: x+(x+y)+y=180∘\text{Substitute angle values: } x + (x + y) + y = 180^\circ
Group like terms: (x+x)+(y+y)=180∘\text{Group like terms: } (x + x) + (y + y) = 180^\circ
2x+2y=180∘2x + 2y = 180^\circ
Factor out 2:2(x+y)=180∘\text{Factor out } 2: 2(x + y) = 180^\circ
Divide both sides by 2:2(x+y)2=180∘2\text{Divide both sides by } 2: \frac{2(x + y)}{2} = \frac{180^\circ}{2}
x+y=90∘x + y = 90^\circ

How to recognise this question in an exam

  • Look for a diameter. It may be stated directly or shown as a line through the centre.
  • Find the angle on the arc. The vertex of the required angle must lie on the semicircle.
  • Show the reasoning before the answer. This earns the method marks instead of presenting a numerical guess.
  • Do not depend on the drawing. Even a stretched or tilted diagram gives the same result.

The useful reverse result

The converse is also useful: if a triangle is right-angled, its hypotenuse is a diameter of the circle through its three vertices. Therefore the midpoint of the hypotenuse is equidistant from all three vertices.

Final step: connect x + y to the required angle

From Step 1, ∠ACB=x+y\angle ACB = x + y. From Step 2, x+y=90∘x + y = 90^\circ.

∠ACB=x+y(From Step 1)\angle ACB = x + y \quad (\text{From Step 1})
x+y=90∘(From Step 2)x + y = 90^\circ \quad (\text{From Step 2})
∠ACB=90∘\boxed{\angle ACB = 90^\circ}

Moving C changes x and y separately, but their sum remains fixed at 90∘90^\circ. That is why the angle subtended by a semicircle is always a right angle (90∘90^\circ), regardless of where point C lies on the circumference.

In Young Math Brains, you identify the diameter, choose the correct arc and justify the theorem before the answer is revealed.

Practise circle theorems step by step

Your turn

If AB were only a chord and did not pass through the centre, would angle ACB still have to be 90 degrees?

Tell us in the comments on YouTube or Instagram, or email connect@youngmathbrains.com. We read every one.

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