Geometric Proof

Both angles stand on the chord AB, and C and D are on the same side of it.

1. Angles ACB and ADB are both at the circumference and both stand on the chord AB
2. C and D are on the same side of AB, so the two angles are in the same segment
3. 'Vertically opposite' needs two angles at the same point, and 'alternate angles' needs parallel lines
4. Answer: Angles in the same segment are equal

The angle y is between a tangent and a chord, and there are no parallel lines in the diagram.

1. Angle TAB is between the tangent TA and the chord AB
2. Angle ACB is in the segment on the other side of AB (the alternate segment)
3. So they are equal by the alternate segment theorem
4. 'Alternate angles' needs parallel lines, which there are not here
5. Answer: Alternate segment theorem

For each option, ask whether the fact quoted is true and whether it applies to the angles named.

1. A tangent is perpendicular to the radius at the point of contact, so angle OAT=90^\circ: correct
2. Opposite angles of a kite do not always add up to 180^\circ, so that reason is not a true fact
3. T is outside the circle, so angle ATB is not an angle at the circumference
4. TA=TB makes triangle TAB isosceles, not triangle OAT
5. (The proof goes on: angle OBT=90^\circ too, so angle AOB=360^\circ-90^\circ-90^\circ-x=180^\circ-x, angles in a quadrilateral.)
6. Answer: Angle OAT=90^\circ because the angle between a tangent and a radius is 90^\circ

What is the size of angle ACB?

1. Angle ACB=90^\circ (the angle in a semicircle is 90^\circ)
2. (3x-4)+(2x+9)+90=180 (angles in a triangle add up to 180^\circ)
3. 5x+95=180, so 5x=85
4. Answer: x=17

For each line, check that the fact quoted is true and that it matches the lines in the diagram.

1. Line 1 is correct: OB=OC (radii), so triangle OBC is isosceles
2. Line 2 states the right angle with the wrong reason: there is no tangent in the diagram
3. Angle ACB=90^\circ because the angle in a semicircle is 90^\circ
4. Line 3 is correct
5. Answer: Line 2

Angles BCD and DAB are opposite angles of the quadrilateral ABCD, whose vertices are all on the circle.

1. ABCD is a cyclic quadrilateral and angles BCD and DAB are opposite angles
2. Opposite angles of a cyclic quadrilateral add up to 180^\circ (they are not equal)
3. AD and BC are not parallel, so co-interior angles do not apply, and neither angle is at the centre
4. Answer: Angle BCD=180^\circ- angle DAB (opposite angles of a cyclic quadrilateral add up to 180^\circ)

Angle ABD stands on the chord AD: find another angle in the diagram that stands on AD.

1. Angle ACD=x (angles in the same segment are equal)
2. In triangle XCD, angle XCD=x and angle XDC=y
3. BXD is a straight line, so angle BXC is an exterior angle of triangle XCD
4. Angle BXC=x+y (the exterior angle of a triangle equals the sum of the two interior opposite angles)
5. Answer: x+y

The angle at the centre that is twice angle ABC is on the other side of O from B.

1. Angle ABC stands on the major arc AC, so the angle at the centre on that arc is the reflex angle AOC
2. Reflex angle AOC=2x (the angle at the centre is twice the angle at the circumference)
3. y+2x=360^\circ (angles around a point add up to 360^\circ)
4. Answer: y=360^\circ-2x

Use the alternate segment theorem to find angle TAB, then use the fact that TA=TB.

1. Angle TAB=x (alternate segment theorem)
2. TA=TB (tangents from an external point are equal in length)
3. So angle TBA=x (base angles of an isosceles triangle are equal)
4. Angle ATB=180^\circ-2x (angles in a triangle add up to 180^\circ)
5. Answer: 180^\circ-2x

OA, OB and OC are all radii, so look for two isosceles triangles.

1. OA=OB (radii), so angle OBA=x (base angles of an isosceles triangle are equal)
2. OC=OB (radii), so angle OBC=y (base angles of an isosceles triangle are equal)
3. So angle ABC=x+y
4. Angle AOC=2(x+y) (the angle at the centre is twice the angle at the circumference)
5. Answer: 2x+2y

Find angle ABC first, then use the fact that ABCD is a cyclic quadrilateral.

1. Angle ABC=180^\circ-x (angles on a straight line add up to 180^\circ)
2. Angle ADC=180^\circ-(180^\circ-x)=x (opposite angles of a cyclic quadrilateral add up to 180^\circ)
3. Triangle ADC is isosceles because DA=DC, so angle DAC= angle DCA (base angles of an isosceles triangle are equal)
4. Angle DAC=\dfrac{180^\circ-x}{2} (angles in a triangle add up to 180^\circ)
5. Answer: 90^\circ-\dfrac{x}{2}

Use the alternate segment theorem to write angle ACB in terms of x.

1. Angle ACB=2x+14^\circ (alternate segment theorem)
2. BA=BC, so angle BAC= angle BCA=2x+14^\circ (base angles of an isosceles triangle are equal)
3. 2(2x+14)+(x+2)=180 (angles in a triangle add up to 180^\circ)
4. 5x+30=180, so 5x=150
5. Answer: x=30

Draw a line through E parallel to AB and look for co-interior angles.

1. Draw the line EF through E parallel to AB and CD, with F on the same side as A and C
2. Angle BEF=180^\circ-x (co-interior angles add up to 180^\circ)
3. Angle DEF=180^\circ-y (co-interior angles add up to 180^\circ)
4. Angle BED=(180^\circ-x)+(180^\circ-y)
5. Answer: 360^\circ-x-y

Start with the exterior angle of the polygon, then use the fact that BA=BC.

1. Each exterior angle is \dfrac{360^\circ}{n}
2. Angle ABC=180^\circ-\dfrac{360^\circ}{n} (an interior and an exterior angle add up to 180^\circ)
3. BA=BC, so angle BAC= angle BCA (base angles of an isosceles triangle are equal)
4. Angle BAC+ angle BCA=180^\circ- angle ABC=\dfrac{360^\circ}{n} (angles in a triangle)
5. Answer: \dfrac{180^\circ}{n}

Use area =\frac{1}{2}ab\sin C in triangle AOB to find angle AOB first.

1. \frac12\times9\times9\times\sin AOB=30
2. \sin AOB=\dfrac{60}{81}, so angle AOB=47.79\ldots^\circ
3. OA=OB (radii), so angle OAB= angle OBA (base angles of an isosceles triangle are equal)
4. Angle OAB=\dfrac{180^\circ-47.79\ldots^\circ}{2}=66.10\ldots^\circ
5. Answer: 66.1^\circ

Radii are equal, and a tangent is perpendicular to the radius at the point of contact.

Part (a)

1. OA=OB (radii), so triangle OAB is isosceles
2. Angle OAB= angle OBA=2x+15^\circ (base angles of an isosceles triangle are equal)
3. Angle OAT=90^\circ (the angle between a tangent and a radius is 90^\circ)
4. So (2x+15)+3x=90
5. 5x=75
6. Answer: x=15

Part (b)

1. Angle TAB=3\times15=45^\circ and angle OAB=2\times15+15=45^\circ
2. Angle ACB=45^\circ (alternate segment theorem)
3. Angle CAB=20^\circ+45^\circ=65^\circ
4. Angle ABC=180^\circ-65^\circ-45^\circ=70^\circ (angles in a triangle add up to 180^\circ)
5. Answer: 70^\circ

Angle DBC stands on the same chord as angle DAC. Answer part (a) before attempting part (b).

Part (a)

1. Angle DBC= angle DAC=3x-20^\circ (angles in the same segment are equal)
2. So angle ABC=(x+15)+(3x-20)=4x-5^\circ
3. In triangle ABC: (2x+10)+(4x-5)+(4x-25)=180 (angles in a triangle add up to 180^\circ)
4. 10x-20=180, so 10x=200
5. Answer: x=20

Part (b)

1. Angle BAD=(2x+10)+(3x-20)=5x-10=90^\circ
2. Angle ABC=4x-5=75^\circ, so AC is not a diameter
3. BD makes a right angle at A, a point on the circle, so BD is a diameter (converse of the angle in a semicircle)
4. Angle ABD= angle ACD is true for any chord AD, so it does not show that BD is a diameter
5. Answer: BD is a diameter, because angle BAD=90^\circ

Start with the isosceles triangle OBD. Answer part (a) before attempting part (b).

Part (a)

1. OB=OD (radii), so angle ODB=2x (base angles of an isosceles triangle are equal)
2. Angle BOD=180^\circ-4x (angles in a triangle add up to 180^\circ)
3. Reflex angle BOD=360^\circ-(180^\circ-4x) (angles around a point add up to 360^\circ)
4. Answer: 180^\circ+4x

Part (b)

1. C is on the minor arc BD, so angle BCD stands on the major arc
2. Reflex angle BOD=2\times angle BCD (the angle at the centre is twice the angle at the circumference)
3. 180+4x=2(7x-10)=14x-20
4. 200=10x
5. Answer: x=20

Part (c)

1. Angle BCD=7\times20-10=130^\circ
2. Angle BAD=180^\circ-130^\circ (opposite angles of a cyclic quadrilateral add up to 180^\circ)
3. Answer: 50^\circ

Start with the isosceles triangle TAB, then use the alternate segment theorem. Answer part (a) before attempting part (b).

Part (a)

1. TA=AB, so angle ABT=x (base angles of an isosceles triangle are equal)
2. Angle TAB=180^\circ-2x (angles in a triangle add up to 180^\circ)
3. Angle ACB= angle TAB (alternate segment theorem)
4. Answer: 180^\circ-2x

Part (b)

1. Angle ABC=180^\circ-x (angles on a straight line add up to 180^\circ)
2. Angle BAC=180^\circ-(180^\circ-x)-(180^\circ-2x) (angles in a triangle add up to 180^\circ)
3. =180^\circ-180^\circ+x-180^\circ+2x
4. Answer: 3x-180^\circ

Part (c)

1. BA=BC, so angle BAC= angle BCA (base angles of an isosceles triangle are equal)
2. 3x-180=180-2x
3. 5x=360
4. Answer: x=72

Write AC^2 in two ways with the cosine rule, using triangles ABC and ADC. Answer part (a) before attempting part (b).

Part (a)

1. Angle ADC=180^\circ-\theta (opposite angles of a cyclic quadrilateral add up to 180^\circ)
2. So \cos ADC=\cos(180^\circ-\theta)=-\cos\theta
3. Triangle ABC: AC^2=3^2+5^2-2\times3\times5\cos\theta=34-30\cos\theta
4. Triangle ADC: AC^2=8^2+5^2+2\times8\times5\cos\theta=89+80\cos\theta
5. 34-30\cos\theta=89+80\cos\theta, so 110\cos\theta=-55
6. Answer: \cos\theta=-\dfrac12

Part (b)

1. AC^2=34-30\times\left(-\dfrac12\right)=49
2. Answer: AC=7 cm

Part (c)

1. \theta=120^\circ and angle ADC=60^\circ
2. Area of triangle ABC=\frac12\times3\times5\times\sin120^\circ=6.495\ldots
3. Area of triangle ADC=\frac12\times8\times5\times\sin60^\circ=17.320\ldots
4. Total =23.815\ldots
5. Answer: 23.8 cm^2