Circle theorem proofs and linked geometry

(a) Let OAC=OCA=x and OBC=OCB=y, using equal radii. Then triangle ABC has angles x, y and x+y. Its angle sum gives 2x+2y=180, so ACB=x+y=90 degrees.

1. Use OA=OC and OB=OC to establish both pairs of equal base angles.
2. Write the full triangle angle sum using x and y.
   $$x+y+(x+y)=180$$
3. Let OAC=OCA=x and OBC=OCB=y, using equal radii. Then triangle ABC has angles x, y and x+y. Its angle sum gives 2x+2y=180, so ACB=x+y=90 degrees.

• M1 Use OA=OC and OB=OC to establish both pairs of equal base angles.
• M1 Write the full triangle angle sum using x and y.
• C1 Correct conclusion with supporting reasoning: Let OAC=OCA=x and OBC=OCB=y, using equal radii. Then triangle ABC has angles x, y and x+y. Its angle sum gives 2x+2y=180, so ACB=x+y=90 degrees.

(b) Each relevant triangle has two radii as equal sides, so it is isosceles.

1. Each relevant triangle has two radii as equal sides, so it is isosceles.

• C1 Correct conclusion with supporting reasoning: Each relevant triangle has two radii as equal sides, so it is isosceles.

(a) OT=OU as radii, OP is common, and OT is perpendicular to PT while OU is perpendicular to PU. Right triangles OTP and OUP are congruent by RHS, so PT=PU.

1. Identify the two right angles from the radius-tangent theorem.
2. Use the equal radii and common hypotenuse OP for RHS congruence.
3. Therefore OT=OU as radii, OP is common, and OT is perpendicular to PT while OU is perpendicular to PU. Right triangles OTP and OUP are congruent by RHS, so PT=PU.

• M1 Identify the two right angles from the radius-tangent theorem.
• M1 Use the equal radii and common hypotenuse OP for RHS congruence.
• C1 Correct conclusion with supporting reasoning: OT=OU as radii, OP is common, and OT is perpendicular to PT while OU is perpendicular to PU. Right triangles OTP and OUP are congruent by RHS, so PT=PU.

(b) $12$ cm

1. Apply Pythagoras in the radius-tangent right triangle.
   $$\sqrt{15^{2}-9^{2}}$$
2. Therefore $12$ cm.

• M1 Apply Pythagoras in the radius-tangent right triangle.
• A1 Correct answer: $12$ cm

(a) OA=OB, AM=BM and OM is common. Triangles OAM and OBM are congruent by SSS. Equal angles OMA and OMB lie on a straight line, so each is 90 degrees.

1. Establish all three corresponding side equalities.
2. The equal adjacent angles total 180 degrees.
   $$2x=180$$
3. Therefore OA=OB, AM=BM and OM is common. Triangles OAM and OBM are congruent by SSS. Equal angles OMA and OMB lie on a straight line, so each is 90 degrees.

• M1 Establish all three corresponding side equalities.
• M1 The equal adjacent angles total 180 degrees.
• C1 Correct conclusion with supporting reasoning: OA=OB, AM=BM and OM is common. Triangles OAM and OBM are congruent by SSS. Equal angles OMA and OMB lie on a straight line, so each is 90 degrees.

(b) $12$ cm

1. Use half the chord as one shorter side in a right triangle.
   $$\sqrt{20^{2}-16^{2}}$$
2. Therefore $12$ cm.

• M1 Use half the chord as one shorter side in a right triangle.
• A1 Correct answer: $12$ cm

(a) Using isosceles triangles $OAC$ and $OBC$ and exterior angles.

1. $OA = OC$ (radii), so triangle $OAC$ is isosceles: angle $OCA =$ angle $OAC = x$.
2. The exterior angle of triangle $OAC$ at $O$ is angle $AOD = x + x = 2x$.
3. Similarly $OB = OC$ gives angle $OCB =$ angle $OBC = y$ and angle $BOD = 2y$.
4. So angle $AOB = 2x + 2y = 2(x + y) = 2 \times$ angle $ACB$.

• M1 Using $OA = OC$ (radii) to get equal base angles $x$.
• M1 Angle $AOD = 2x$ (exterior angle, or $180 - (180 - 2x)$).
• M1 The same argument for triangle $OBC$, giving $2y$.
• C1 A complete argument reaching angle $AOB = 2(x + y) = 2 \times$ angle $ACB$, with every step justified.

(a) Let OCA=x and OCB=y. Equal radii give OAC=x and OBC=y. Angles AOC and BOC are 180-2x and 180-2y. Angles around O then give minor AOB=2x+2y=2ACB.

1. Use two isosceles triangles formed by equal radii.
2. Subtract their central angles from a full turn.
   $$360-(180-2x)-(180-2y)=2x+2y$$
3. Let OCA=x and OCB=y. Equal radii give OAC=x and OBC=y. Angles AOC and BOC are 180-2x and 180-2y. Angles around O then give minor AOB=2x+2y=2ACB.

• M1 Use two isosceles triangles formed by equal radii.
• M1 Subtract their central angles from a full turn.
• C1 Correct conclusion with supporting reasoning: Let OCA=x and OCB=y. Equal radii give OAC=x and OBC=y. Angles AOC and BOC are 180-2x and 180-2y. Angles around O then give minor AOB=2x+2y=2ACB.

(b) This fixes which angles add and which central angle is minor; other diagrams require different angle decompositions.

1. This fixes which angles add and which central angle is minor; other diagrams require different angle decompositions.

• C1 Correct conclusion with supporting reasoning: This fixes which angles add and which central angle is minor; other diagrams require different angle decompositions.

(a) The two arcs AC together form the full circle. The two central angles subtending those arcs sum to 360 degrees. Each corresponding circumference angle is half its central angle, so ABC+ADC=180 degrees.

1. Identify the two complementary arcs AC and their full-turn central total.
   $$x+y=360$$
2. Apply the centre-to-circumference angle relation to both arcs.
   $$x/2+y/2=180$$
3. The two arcs AC together form the full circle. The two central angles subtending those arcs sum to 360 degrees. Each corresponding circumference angle is half its central angle, so ABC+ADC=180 degrees.

• M1 Identify the two complementary arcs AC and their full-turn central total.
• M1 Apply the centre-to-circumference angle relation to both arcs.
• C1 Correct conclusion with supporting reasoning: The two arcs AC together form the full circle. The two central angles subtending those arcs sum to 360 degrees. Each corresponding circumference angle is half its central angle, so ABC+ADC=180 degrees.

(b) Then B and D lie on opposite arcs AC, so the two angles subtend the complementary arcs used in the proof.

1. Then B and D lie on opposite arcs AC, so the two angles subtend the complementary arcs used in the proof.

• C1 Correct conclusion with supporting reasoning: Then B and D lie on opposite arcs AC, so the two angles subtend the complementary arcs used in the proof.

(a) Angle $TAB = 90 -$ angle $DAB =$ angle $ADB =$ angle $ACB$.

1. Angle $TAD = 90^\circ$: the tangent is perpendicular to the radius (diameter) at $A$.
2. So angle $TAB = 90^\circ -$ angle $BAD$.
3. Angle $ABD = 90^\circ$: the angle in a semicircle ($AD$ is a diameter).
4. In triangle $ABD$: angle $ADB = 180 - 90 -$ angle $BAD = 90^\circ -$ angle $BAD$. So angle $TAB$ = angle $ADB$.
5. Angle $ADB$ = angle $ACB$: angles in the same segment (both stand on the chord $AB$).
6. So angle $TAB$ = angle $ACB$.

• M1 Angle $TAB = 90^\circ -$ angle $BAD$ from tangent perpendicular to the radius.
• M1 Angle $ADB = 90^\circ -$ angle $BAD$ using the angle in a semicircle.
• M1 Angle $ADB$ = angle $ACB$ (angles in the same segment).
• C1 A complete chain with every reason stated.

(a) $\frac{240}{13}$ cm

1. $$\sqrt{26^{2}-10^{2}}$$
2. $$2\times 10\times 24/26$$
3. OT is perpendicular to PT. By Pythagoras, PT = $\sqrt{}$($26^{2}$ $-$ $10^{2}$) = 24 cm; PU is also 24 cm.
4. The area of quadrilateral OTPU is the sum of two right triangles: 2 $\times$ $\frac{1}{2}$ $\times$ 10 $\times$ 24 = 240 cm².
5. Equal radii and equal tangents make OTPU a kite. Its diagonals OP and TU are perpendicular, so its area is $\frac{1}{2}$ $\times$ OP $\times$ TU = 13 $\times$ TU.
6. Therefore TU = 240/13 cm.

• P1 Establishing $\sqrt{26^{2}-10^{2}}$ or an equivalent valid method.
• P1 Establishing $2\times 10\times 24/26$ or an equivalent valid method.
• A1 Correct answer: $\frac{240}{13}$ cm