Circle Theorems (GCSE Maths)

• A tangent to a circle forms a $90^\circ$ angle with the radius at the point of contact.
• Therefore, $\angle OCD = 90^\circ$.
• To find $\angle ACO$, subtract $\angle ACD$ from the full $90^\circ$ angle.

• By the alternate segment theorem, the angle between the tangent and a chord through the point of contact is equal to the angle in the alternate segment.
• Therefore, $\angle ABC$ is equal to $\angle ACD$.

• Triangle $ABC$ is an isosceles triangle because two tangents drawn from the same external point ($B$) to a circle are equal in length ($AB = BC$).
• Angles at the base of an isosceles triangle are equal.

A tangent meets a radius at exactly $90^\circ$.

Angle $OAC$ is the difference between the full $90^\circ$ angle and angle $CAB$.

• Points $A$, $B$, $C$, and $D$ form a cyclic quadrilateral because all four points lie on the circumference of the circle.
• This means that $\angle ABC$ and $\angle ADC$ are opposite angles in this quadrilateral.

Subtract $\angle ADC$ from $180^\circ$ to find the missing opposite angle:

The required geometric reason relates to the properties of the shape identified in the previous part.

The geometric reason is that angles in the same segment are equal.

• The angle subtended by an arc at the centre is double the angle subtended by it at any remaining part of the circle.
• Alternatively, the angle at the circumference is half the angle at the centre.

The tangent $DA$ meets radius $OA$ at $90^\circ$, creating a right-angled triangle $OAD$.

By symmetry, the angle $AOC$ is twice the angle $AOD$:

Since angles around a point sum to $360^\circ$:

Using the arc length formula $\dfrac{\theta}{360^\circ} \times 2\pi r$ with $r = 5\text{ cm}$:

• Lines $OA$ and $OB$ are radii of the circle, forming an isosceles triangle ($AOB$).
• Base angles are equal, so $\angle BAO = x^\circ$.

• $BC$ is a tangent and $OB$ is a radius, meeting at $90^\circ$.
• $\angle OBC = 90^\circ$.

The angle $ABC$ is the sum of angles $\angle ABO$ and $\angle OBC$:

In triangle $ABC$, the sum of angles is $180^\circ$:

• A tangent meets a radius at exactly $90^\circ$, so the total angle $OAE$ is $90^\circ$.

• Triangle $AOB$ is an isosceles triangle because $OA$ and $OB$ are both radii of the same circle.
• Base angles of an isosceles triangle are equal, so angle $OBA = \angle OAB = 34^\circ$.

• By the Alternate Segment Theorem, the angle between the tangent $DAE$ and chord $AC$ is equal to the angle in the alternate segment ($\angle CBA$).

The total angle between the radius $OA$ and the tangent $DAE$ is $90^\circ$ ($\angle OAD = 90^\circ$).

• Triangle $ABD$ is an isosceles triangle because we are given that $BA = BD$.
• Base angles of an isosceles triangle are equal, and angles in a triangle add up to $180^\circ$.

(Note: $\angle BAD$ is also $70^\circ$)

• $ABCD$ is a cyclic quadrilateral because all four points lie on the circumference of the circle.
• Opposite angles in a cyclic quadrilateral add up to $180^\circ$.

• Triangle $BCD$ is an isosceles triangle because we are given that $CB = CD$.
• Base angles of an isosceles triangle are equal, and angles in a triangle add up to $180^\circ$.

• We are given that $CDE$ is a straight line. Angles on a straight line add up to $180^\circ$.
• The angles around point $D$ on this line are $\angle BDC$, $\angle BDA$, and $\angle ADE$.

A tangent to a circle meets the radius at a right angle ($90^\circ$).

The sum of the interior angles in any triangle is always $180^\circ$.

The points $S$, $O$, and $P$ form a straight line, and angles on a straight line add up to $180^\circ$.

• The lines $OS$ and $OT$ are both radii of the same circle, meaning they are equal in length.
• This makes triangle $SOT$ an isosceles triangle, so the base angles are equal ($\angle OST = \angle OTS = x$).
• The sum of the angles in triangle $SOT$ is $180^\circ$.

Consider the triangles $\triangle OAB$ and $\triangle OCB$.

• $OA = OC$: Both are radii of the same circle.
• $AB = CB$: Given in the problem.
• $OB = OB$: Common side shared by both triangles.

Since all three sides are equal, the triangles are congruent by the Side-Side-Side (SSS) congruence theorem:

Since the triangles are congruent, their corresponding angles must be equal:

• $ABCD$ is a cyclic quadrilateral, so opposite angles sum to $180^\circ$.
• Let $\angle BCD = y$. Given the ratio $3:1$, $\angle BAD = 3y$.

In triangle $ABD$, the sum of internal angles is $180^\circ$.

The Alternate Segment Theorem states that the angle between the tangent $SBT$ and the chord $AB$ is equal to the angle subtended by the chord in the alternate segment.

• The angle at the centre of a circle is twice the angle at the circumference. Therefore, $\angle BAD = \frac{132^\circ}{2} = 66^\circ$.
• In a cyclic quadrilateral, opposite angles sum to $180^\circ$, so $\angle BCD = 180^\circ - 66^\circ = 114^\circ$.

Since $BCE$ is a straight line, the exterior angle $\angle DCE$ is supplementary to the interior angle $\angle BCD$:

In triangle $CDE$, the sum of interior angles is $180^\circ$:

• The angle between a tangent ($CD$) and a chord ($BC$) is equal to the angle subtended by the chord in the alternate segment.
• Therefore, $\angle BAC = \angle BCD = 40^\circ$.

• In $\triangle OAB$, $OA = OB$ (radii). Thus, $\angle OBA = \angle OAB$.
• Let $\angle OAC = y$. Then $\angle OAB = 3y$.
• Since $\angle CAB = 2y$ and $2y = 40^\circ$, it follows that $y = 20^\circ$.
• Thus, $\angle OAC = 20^\circ$ and $\angle OAB = 60^\circ$.

• Since $OA = OC$ (radii), $\triangle OAC$ is isosceles.
• Therefore, $\angle OCA = \angle OAC = 20^\circ$.

We know $\angle OCD = 90^\circ$ because the tangent meets the radius at $90^\circ$.

Since $OA$ and $OB$ are radii of the circle, $OA = OB$.

• Therefore, triangle $AOB$ is an isosceles triangle.
• The base angles are equal because radii of the same circle are equal in length.

The sum of angles in a triangle is $180^\circ$. Since the base angles are equal:

The tangent $DBC$ is perpendicular to the radius $OB$ at point $B$:

Calculate $\angle ABC$ as the difference between $\angle OBC$ and $\angle OBA$: