Circle Theorems (GCSE Maths)

Step 1: Find angle OAB

A tangent to a circle meets the radius at exactly 90°. Since $DAE$ is a tangent and $OA$ is a radius, the total angle $OAE$ is 90°.

We are given that angle $BAE$ is 56°.

• Angle $OAB$ = 90° - 56° = 34°

Step 2: Find the total angle CBA

Look at triangle $AOB$. The lines $OA$ and $OB$ are both radii of the circle, meaning they are the same length. This makes triangle $AOB$ an isosceles triangle, so its base angles must be equal:

• Angle $OBA$ = Angle $OAB$ = 34°

We are given that angle $CBO$ is 35°. Now we can find the total angle for $CBA$:

• Angle $CBA$ = Angle $CBO$ + Angle $OBA$

• Angle $CBA$ = 35° + 34° = 69°

Step 3: Find angle CAD using the Alternate Segment Theorem

The Alternate Segment Theorem states that the angle between a tangent and a chord is equal to the angle in the alternate segment.

• The angle between tangent $DAE$ and chord $AC$ is angle $CAD$.

• The angle in the alternate segment is angle $CBA$ (which we just calculated as 69°).

• Therefore, angle $CAD$ = 69°

Step 4: Calculate angle CAO

We know that the total angle between the radius $OA$ and the tangent $DAE$ is 90° (angle $OAD$ = 90°).

This 90° angle is made up of angle $CAO$ and angle $CAD$.

• Angle $OAD$ = Angle $CAO$ + Angle $CAD$

• 90° = Angle $CAO$ + 69°

• Angle $CAO$ = 90° - 69° = 21°

Step 1: Find angle $BAO$

Lines $OA$ and $OB$ are both radii of the circle, making triangle $AOB$ an isosceles triangle.

Therefore, angle $BAO = x^{\circ}$ because base angles in an isosceles triangle are equal.

Step 2: Find angle $OBC$

The line $BC$ is a tangent to the circle at point $B$, and $OB$ is the radius.

Therefore, angle $OBC = 90^{\circ}$ because a tangent to a circle is perpendicular to the radius.

Step 3: Define the total angle $ABC$

Angle $ABC$ is the sum of angle $ABO$ and angle $OBC$.

Angle $ABC = x^{\circ} + 90^{\circ}$.

Step 4: Calculate angle $ACB$ using the large triangle $ABC$

In the large triangle $ABC$, the sum of the internal angles must be $180^{\circ}$ because angles in a triangle add up to $180^{\circ}$.

Angle $BAO$ + Angle $ABC$ + Angle $ACB = 180^{\circ}$.

Substitute the known values into the equation: $x^{\circ} + (x^{\circ} + 90^{\circ}) + \text{Angle } ACB = 180^{\circ}$.

Combine the like terms: $2x^{\circ} + 90^{\circ} + \text{Angle } ACB = 180^{\circ}$.

Step 5: Simplify to find the final answer

Rearrange the equation to solve for Angle $ACB$:

Angle $ACB = 180^{\circ} - 90^{\circ} - 2x^{\circ}$

Angle $ACB =$ $90 - 2x$

Step 1: Find angle $OAD$.

The line $CAD$ is a tangent to the circle at point $A$, and $OA$ is the radius. A tangent is always perpendicular ($90^\circ$) to the radius at the point of contact.

• Angle $OAD = 90^\circ$

Step 2: Find angle $AOD$.

Look at the right-angled triangle $OAD$. The angles in any triangle add up to $180^\circ$.

• Angle $AOD = 180^\circ - 90^\circ - 32^\circ$

• Angle $AOD = 58^\circ$

Step 3: Find angle $OAB$.

Because $BOD$ is a straight line, angles on this line add up to $180^\circ$.

• Angle $AOB = 180^\circ - 58^\circ = 122^\circ$

Now, look at triangle $AOB$. Because $OA$ and $OB$ are both radii of the same circle, they are equal in length. This makes triangle $AOB$ an isosceles triangle, meaning its base angles ($OAB$ and $OBA$) are equal.

• Angle $OAB = \frac{180^\circ - 122^\circ}{2}$

• Angle $OAB = \frac{58^\circ}{2} = 29^\circ$

(Shortcut: You can also use the exterior angle theorem here. The exterior angle $AOD$ ($58^\circ$) is equal to the sum of the opposite interior angles $OAB$ and $OBA$. Since they are equal, $58^\circ \div 2 = 29^\circ$.)

Step 4: Calculate angle $CAB$.

Just like angle $OAD$, the angle $OAC$ on the other side of the radius is also $90^\circ$ because $CAD$ is a straight tangent line.

Angle $OAC$ is made up of angle $CAB$ and angle $OAB$.

• Angle $CAB = \text{Angle } OAC - \text{Angle } OAB$

• Angle $CAB = 90^\circ - 29^\circ$

• Angle $CAB = 61^\circ$

Step 1: Identify angles from the equilateral triangle

Because triangle $AED$ is an equilateral triangle, all of its interior angles are $60^\circ$.

Specifically, angle $DAE = 60^\circ$ and angle $ADE = 60^\circ$.

Since $AEC$ and $DEB$ are straight lines, we can also refer to these angles as angle $DAC = 60^\circ$ and angle $ADB = 60^\circ$.

Step 2: Find corresponding equal angles using circle theorems

Using the circle theorem that angles in the same segment are equal:

• Angle $ACB$ and angle $ADB$ are both subtended by the same arc ($AB$).

  Therefore, angle $ACB$ = angle $ADB = 60^\circ$.

• Angle $DBC$ and angle $DAC$ are both subtended by the same arc ($DC$).

  Therefore, angle $DBC$ = angle $DAC = 60^\circ$.

From this, we have proven that angle $ACB$ = angle $DBC$.

Step 3: Identify a second pair of equal angles

Using the same circle theorem (angles in the same segment are equal), look at the angles subtended by arc $BC$:

• angle $BAC$ = angle $BDC$

Step 4: Identify a common side

Both triangle $ABC$ and triangle $DCB$ share the base line $BC$.

• $BC$ is a common side ($BC = CB$).

Step 5: Conclude congruency

Now, let's look at the two triangles, $\triangle ABC$ and $\triangle DCB$, with the facts we have proven:

1. Angle $BAC$ = Angle $BDC$ (Angle)

2. Angle $ACB$ = Angle $DBC$ (Angle)

3. Side $BC$ = Side $CB$ (Side)

Because we have two matching angles and a corresponding matching side, we can conclude that triangle $ABC$ is congruent to triangle $DCB$ based on the AAS (Angle-Angle-Side) condition.

Step 1: Find angle $BDA$

Look at triangle $ABD$. We are given that $BA = BD$, which means triangle $ABD$ is an isosceles triangle.

• Reason: Base angles of an isosceles triangle are equal.

• Reason: Angles in a triangle add up to $180^\circ$.

• Angle $BAD$ = Angle $BDA$ = $\frac{180^\circ - 40^\circ}{2}$

• Angle $BDA = \frac{140^\circ}{2} = 70^\circ$

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

Step 2: Find angle $BCD$

The points $A$, $B$, $C$, and $D$ all lie on the circumference of the circle, making $ABCD$ a cyclic quadrilateral.

• Reason: Opposite angles of a cyclic quadrilateral add up to $180^\circ$.

• Angle $BCD$ + Angle $BAD = 180^\circ$

• Angle $BCD + 70^\circ = 180^\circ$

• Angle $BCD = 180^\circ - 70^\circ = 110^\circ$

Step 3: Find angle $BDC$

Look at triangle $BCD$. We are given that $CB = CD$, which means triangle $BCD$ is also an isosceles triangle.

• Reason: Base angles of an isosceles triangle are equal.

• Reason: Angles in a triangle add up to $180^\circ$.

• Angle $CBD$ = Angle $BDC$ = $\frac{180^\circ - 110^\circ}{2}$

• Angle $BDC = \frac{70^\circ}{2} = 35^\circ$

Step 4: Find angle $ADE$

We are given that $CDE$ is a straight line.

• Reason: Angles on a straight line add up to $180^\circ$.

  The angles on the straight line around point $D$ are angle $BDC$, angle $BDA$, and angle $ADE$.

• Angle $BDC$ + Angle $BDA$ + Angle $ADE = 180^\circ$

• $35^\circ + 70^\circ + \text{Angle } ADE = 180^\circ$

• $105^\circ + \text{Angle } ADE = 180^\circ$

• Angle $ADE = 180^\circ - 105^\circ$

• Angle $ADE = 75^\circ$

Step 1: Add a construction line to the diagram.

Draw a straight line from the center of the circle, $O$, to point $C$ on the circumference. This divides the large triangle $ABC$ into two smaller triangles: $\triangle AOC$ and $\triangle BOC$.

Step 2: Identify isosceles triangles and assign variables.

The lines $OA$, $OB$, and $OC$ all represent the radius of the same circle, meaning they are all equal in length.

Because $OA = OC$, triangle $AOC$ is an isosceles triangle. Therefore, its base angles are equal. Let's call these angles $x$:

• Angle $OAC = x$

• Angle $OCA = x$

Because $OB = OC$, triangle $BOC$ is also an isosceles triangle. Its base angles are also equal. Let's call these angles $y$:

• Angle $OBC = y$

• Angle $OCB = y$

Step 3: Define the angles of the large triangle.

Look at the large triangle $\triangle ABC$. Its three interior angles can now be written in terms of $x$ and $y$:

• Angle at $A$ (Angle $BAC$) = $x$

• Angle at $B$ (Angle $ABC$) = $y$

• Angle at $C$ (Angle $ACB$) = Angle $OCA$ + Angle $OCB$ = $x + y$

Step 4: Use the property of angles in a triangle.

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

Applying this to the large triangle $\triangle ABC$:

Step 5: Solve the equation to prove the angle.

Combine the like terms in the equation:

Divide the entire equation by $2$:

Since angle $ACB$ is exactly equal to $x + y$, we have proven that angle $ACB = 90^\circ$.

Step 1: Identify the right-angled triangles.

A tangent to a circle forms a $90^\circ$ angle with the radius at the point of contact. Therefore, angle $OAD$ and angle $OCD$ are both $90^\circ$. This means triangle $OAD$ is a right-angled triangle.

Step 2: Find angle $AOD$ using trigonometry.

In the right-angled triangle $OAD$:

• The adjacent side to angle $AOD$ is the radius $OA = 5\text{ cm}$.

• The hypotenuse is $DO = 9\text{ cm}$.

Using the cosine ratio ($\cos \theta = \frac{\text{Adjacent}}{\text{Hypotenuse}}$):

Step 3: Find the angle for the major arc $ABC$.

Due to symmetry, triangle $OAD$ is congruent to triangle $OCD$, so angle $COD$ is also $56.251^\circ$.

The minor angle $AOC = 2 \times 56.251^\circ = 112.502^\circ$.

The arc $ABC$ is the major arc, which corresponds to the reflex angle $AOC$.

Angles around a point add up to $360^\circ$, so:

Step 4: Calculate the length of arc $ABC$.

The formula for arc length is $\frac{\theta}{360^\circ} \times 2\pi r$, where $\theta$ is the central angle and $r$ is the radius.

Step 5: Round the answer.

Rounding to 3 significant figures gives:

$21.6\text{ cm}$

Step 1: Find the first pair of equal angles using circle theorems.

Look at the angles subtended by arc $BC$.

• Angle $BAE$ (which is the same as angle $BAC$) and angle $CDE$ (which is the same as angle $BDC$) are both subtended by the same arc $BC$.

• Therefore, angle $BAE$ = angle $CDE$.

• Reason: Angles in the same segment are equal.

Step 2: Find a second pair of equal angles at the intersection.

Look at the point $E$ where the straight lines $AEC$ and $BED$ cross.

• Angle $AEB$ = angle $DEC$.

• Reason: Vertically opposite angles are equal.

Step 3: Find the third pair of equal angles.

You can find the third pair using another circle theorem or the angle sum of a triangle.

• Using arc $AD$, angle $ABE$ (same as $ABD$) and angle $DCE$ (same as $ACD$) are both subtended by arc $AD$.

• Therefore, angle $ABE$ = angle $DCE$.

• Reason: Angles in the same segment are equal. (Alternatively, since the angles in any triangle add up to $180^\circ$, and two pairs of angles are already proven equal, the third pair must also be equal).

Conclusion:

Because all three corresponding angles in triangle $ABE$ and triangle $DCE$ are equal, the two triangles are similar.

Step 1: Find angle OTP

A tangent to a circle meets the radius at a right angle (90°). Since PT is the tangent and OT is the radius, the angle between them is 90°.

• Angle OTP = 90°

Step 2: Find angle TOP

Look at the right-angled triangle OTP. The sum of the interior angles in any triangle is always 180°.

• Angle TOP = 180° - (Angle OTP + Angle OPT)

• Angle TOP = 180° - (90° + 32°)

• Angle TOP = 180° - 122°

• Angle TOP = 58°

Step 3: Find angle SOT

The points S, O, and P form a straight line. Angles on a straight line add up to 180°.

• Angle SOT = 180° - Angle TOP

• Angle SOT = 180° - 58°

• Angle SOT = 122°

Step 4: Find angle x (Angle OTS)

Look at triangle SOT. 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.

Because it is an isosceles triangle, the base angles are equal: Angle OST = Angle OTS = $x$.

The sum of the angles in triangle SOT is 180°.

• $x + x + \text{Angle SOT} = 180°$

• $2x + 122° = 180°$

• $2x = 180° - 122°$

• $2x = 58°$

• $x = 29°$

(Alternative shortcut for Steps 3 & 4: The exterior angle of a triangle is equal to the sum of the two opposite interior angles. For triangle SOT, the exterior angle is TOP (58°). Therefore, $x + x = 58°$, which means $2x = 58°$, so $x = 29°$.)