Circle Theorems: Question 1

Worked solution

Setting up

All four points lie on the circle, $O$ is the centre, and $AC$ is a diameter because $AOC$ is a straight line through $O$. We will use two key circle theorems:

• Angle in a semicircle: the angle subtended by a diameter at the circumference is $90^\circ$.
• Angle at the centre: the angle subtended by an arc at the centre is twice the angle subtended by the same arc at the circumference.

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(a) Finding $\angle ABC$

$B$ is a point on the circle, and $A$ and $C$ are the two ends of the diameter $AC$. Therefore $\angle ABC$ is the angle in a semicircle.

$\boxed{\angle ABC = 90^\circ}$

Reason: the angle in a semicircle is $90^\circ$ (equivalently, the angle subtended by a diameter at the circumference is a right angle).

Note: the value $35^\circ$ is not needed here; the answer is $90^\circ$ for any position of $B$ on the circle.

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(b) Finding $\angle BCA$

Look at triangle $ABC$. We now know two of its angles:

$\angle BAC = 35^\circ, \qquad \angle ABC = 90^\circ.$

The angles of a triangle sum to $180^\circ$:

$\angle BCA = 180^\circ - 90^\circ - 35^\circ = \boxed{55^\circ}$

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(c) Finding $\angle DBC$

Consider the arc $DC$ (the minor arc, on the side away from $B$).

• At the centre, this arc subtends $\angle DOC = 80^\circ$.
• At the circumference, the same arc subtends $\angle DBC$, because $B$ lies on the major arc $DC$.

By the theorem the angle at the centre is twice the angle at the circumference:

$\angle DOC = 2 \times \angle DBC$

$80^\circ = 2 \times \angle DBC \quad\Longrightarrow\quad \angle DBC = \frac{80^\circ}{2} = \boxed{40^\circ}$

Circle theorem used: the angle subtended by an arc at the centre is twice the angle subtended at the circumference.

Alternative: Point $A$ also lies on the major arc $DC$, so $\angle DAC$ subtends the same arc $DC$. The same theorem gives $\angle DAC = \tfrac{1}{2}\times 80^\circ = 40^\circ$, and $\angle DBC = \angle DAC = 40^\circ$ by angles in the same segment, a useful check.

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(d) Finding $\angle OCD$

In triangle $OCD$, the sides $OC$ and $OD$ are both radii of the circle, so $OC = OD$ and the triangle is isosceles. Hence the two base angles are equal:

$\angle OCD = \angle ODC.$

The angles of triangle $OCD$ sum to $180^\circ$, and $\angle DOC = 80^\circ$, so

$\angle OCD = \frac{180^\circ - 80^\circ}{2} = \frac{100^\circ}{2} = \boxed{50^\circ}$

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Summary of answers