Proving Circle Theorems

Match the three stated equalities to a congruence condition.

1. OA=OB because they are radii, AM=BM because M is the midpoint, and OM is shared.
2. The three pairs of equal sides prove the triangles congruent by SSS.

Use the position of A, M and B as well as the equality.

1. A, M and B lie on a straight line, so \angle OMA+\angle OMB=180^\circ (angles on a straight line).
2. The two angles are equal, so each one is 180^\circ\div2=90^\circ. So OM is perpendicular to AB.
3. The other statements don't prove it: not every line from O to AB is perpendicular to it, and the angle sum of triangle OMA alone doesn't fix the size of \angle OMA.

Identify the hypotenuse and one shared side.

1. OM\perp AB, so each of triangles OMA and OMB has a right angle at M.
2. OA=OB because both are radii: these are the hypotenuses. OM is a side of both triangles.
3. Right angle, hypotenuse and one side: the triangles are congruent by RHS, so the corresponding sides AM and MB are equal.
4. SAS needs the angle between the two sides, but the right angle is not between OA and OM. AAA only shows the triangles are similar. Saying M is the midpoint assumes what is being proved.

Use half of each equal chord as one pair of matching sides.

1. M and N are the midpoints, so AM=\frac12AB and CN=\frac12CD. The chords are equal, so AM=CN.
2. OA=OC because both are radii.
3. Triangles OMA and ONC each have a right angle, equal hypotenuses OA and OC, and equal sides AM and CN, so they are congruent by RHS.
4. So the corresponding sides OM and ON are equal.
5. RHS needs a pair of equal sides as well as the hypotenuses, and SAS needs two pairs of sides. Perpendiculars from the centre are only equal for equal chords, which is what is being proved.

Use the two isosceles triangles and the angles around O.

1. OA=OC (radii), so triangle OAC is isosceles: \angle OAC=x^\circ and \angle AOC=(180-2x)^\circ.
2. Similarly OB=OC, so \angle BOC=(180-2y)^\circ.
3. Angles around O add to 360^\circ: \angle AOB=360-(180-2x)-(180-2y)=2x+2y.
4. The radius ticks justify the equal base angles. The centre is split into \alpha=180^\circ-2x, \beta=180^\circ-2y and the required angle \theta.

5. So \angle AOB=(2x+2y)^\circ, which is twice \angle ACB=(x+y)^\circ. This proves that the angle at the centre is twice the angle at the circumference.

Use the right angle between the tangent and the radius OA, then the isosceles triangle OAB.

1. A tangent is perpendicular to the radius at the point of contact, so \angle OAE=90^\circ.
2. \angle OAB=\angle OAE-\angle BAE=(90-x)^\circ.
3. OA=OB because both are radii, so triangle OAB is isosceles and \angle OBA=(90-x)^\circ too.
4. Angles in triangle OAB add to 180^\circ: \angle AOB=180-2(90-x)=2x, so \angle AOB=(2x)^\circ.
5. The tangent right angle leaves \alpha=90^\circ-x at A. Equal radii give the same angle at B, so the centre angle is 2x.

6. This proves that the angle at the centre is twice the angle between the tangent and the chord AB.

Apply the centre-angle theorem to each vertex.

1. The angle at the centre on arc AB is twice \angle ACB.
2. The same centre angle is twice \angle ADB.
3. Both circumference angles are therefore half of the same angle, proving they are equal.
4. The non-reflex angle AOB is twice each angle, not equal to it. C and D are on the same side of AB, so they are not opposite corners of a cyclic quadrilateral.

Add the two halves of the centre angles.

1. The opposite circumference angles stand on the two arcs AC.
2. Each is half its corresponding centre angle, so their sum is half of 360^\circ.
3. The two opposite angles use the two different arcs from A to C. Adding their halves gives half a full turn.

4. Therefore opposite angles of a cyclic quadrilateral add to 180^\circ.

Use a straight angle at the centre.

1. AC is a diameter, so AOC is a straight line and the angle at the centre is \angle AOC=180^\circ.
2. \angle ABC stands on the same arc AC, and the angle at the centre is twice the angle at the circumference.
3. So \angle ABC=180^\circ\div2=90^\circ.
4. An angle at the circumference is half the angle at the centre, not equal to it. Every triangle's angles add to 180^\circ, so that fact alone can't show one angle is 90^\circ.

Look for the same side in both right-angled triangles.

1. Each tangent is perpendicular to its radius, and OA=OB because both are radii.
2. OP belongs to both triangles and is the hypotenuse in each.
3. RHS congruence follows, so corresponding tangent lengths PA and PB are equal.
4. AB is not a side of either triangle, and \angle APB is not 90^\circ in general. Using AP=BP would assume the result being proved.

The two angles must have their vertex at P.

1. The matching vertices in the congruent triangles are O with O, A with B, and P with P.
2. Thus \angle APO=\angle OPB. These are the two parts of \angle APB cut by OP.
3. Therefore OP bisects the angle at P.
4. OA=OB and \angle OAP=\angle OBP are true, but they are about the sides and the angles at A and B, not the angle at P.

M lies on the angle bisector already established.

1. Triangles OAP and OBP are congruent by RHS, so \angle AOP=\angle BOP. M lies on OP, so \angle AOM=\angle BOM.
2. In triangles OMA and OMB: OA=OB (radii), OM is common, and the angles between them are equal. That is SAS congruence.
3. Only the radii, the shared side OM and the established equal angles are used for SAS. The right angles at M follow after this congruence.

4. So \angle OMA=\angle OMB. They lie on the straight line AB, so both are 90^\circ. Hence OP\perp AB.
5. SSS and RHS would use AM=MB or the right angles at M, but those are what this proof goes on to show, so they can't be used yet.

The tangent is perpendicular to the diameter at A.

1. The tangent is perpendicular to the diameter at A, so \angle TAC=90^\circ and \angle TAB=(90-x)^\circ.
2. \angle ABC=90^\circ because AC is a diameter, so the angles of triangle ABC give \angle ACB=(90-x)^\circ.
3. The diameter gives a right angle at B, so the remaining angle at C matches the tangent angle 90^\circ-x.

4. So \angle TAB=\angle ACB: this is the alternate segment theorem for the chord AB.

Combine the given angle equality with the right angle in triangle ABC.

1. \angle ABC=90^\circ because it stands on the diameter AC, so in triangle ABC: \angle ACB+\angle BAC=90^\circ.
2. \angle TAB=\angle ACB, so \angle TAB+\angle BAC=90^\circ. That sum is \angle TAC, and O lies on AC, so \angle TAO=90^\circ.
3. Let x=\angle BAC, and mark the two equal angles \angle TAB and \angle ACB as \theta. The right angle at B makes \theta+x=90^\circ, and \theta+x is the whole of \angle TAO.

4. A line through A perpendicular to the radius OA is the tangent at A.
5. Quoting the alternate segment theorem here gets it backwards: that theorem starts from a tangent, so using it assumes the result being proved.

An angle bisector divides an angle at its vertex into two equal parts.

1. The equal sides are OA=OB, AM=BM and common side OM.
2. SSS congruence gives \angle AOM=\angle MOB. These are precisely the two angles into which OM divides \angle AOB.
3. Therefore OM bisects the centre angle.
4. \angle OMA=\angle OMB is also true, but those angles are at M, not at O.

Find the angle at the centre in terms of x, then use quadrilateral OAPB.

1. The angle at the centre is twice the angle at the circumference, so \angle AOB=(2x)^\circ.
2. A tangent meets the radius at 90^\circ, so \angle OAP=\angle OBP=90^\circ.
3. The angles of quadrilateral OAPB add to 360^\circ: \angle APB=360-90-90-2x=180-2x.
4. The blue centre angle is twice x. The two tangent-contact right angles leave the red angle \theta=180^\circ-2x at P.

5. So \angle APB=(180-2x)^\circ. (Another way: by the alternate segment theorem \angle PAB=\angle PBA=x^\circ, and the angles of triangle PAB give the same answer.)

Relate the equal circumference angles to triangles made with radii.

1. \angle BAC=\angle ACD by alternate angles because AB\parallel CD.
2. The parallel sides give the two angles marked \theta. Each of these angles stands on one of the chords being compared.

3. The centre angles on chords BC and AD are twice these equal angles, so they are equal.
4. Joining the endpoints of each chord to the centre gives two triangles with two equal radii and equal included angles.
5. At the centre O, \angle BOC=\angle AOD=2\theta. With OB=OA and OC=OD (radii), the two shaded triangles satisfy SAS.

6. SAS congruence gives BC=AD.
7. A cyclic quadrilateral with one pair of parallel sides is a trapezium, not a parallelogram. SAS needs two pairs of equal sides, but only AC is known to be shared.

Write each angle in terms of the interior angle at C.

1. The straight line gives \angle BCE=180^\circ-\angle BCD.
2. The cyclic quadrilateral gives \angle BAD=180^\circ-\angle BCD.
3. The expressions are equal, so an exterior angle of a cyclic quadrilateral equals the opposite interior angle.
4. Opposite angles of a cyclic quadrilateral add to 180^\circ; they are not equal. Alternate angles need parallel lines, which we are not given.

Identify the chord used by both circumference angles.

1. Because AC is a diameter, \angle ABC=90^\circ. Triangle angles and the perpendicular tangent give \angle TAB=\angle ACB.
2. C and D are in the same segment of chord AB, so \angle ACB=\angle ADB.
3. Combining these equalities proves the alternate segment theorem for the angle at D.
4. C and D are on the same side of AB, so the angles at C and D are not opposite angles of a cyclic quadrilateral. Only \angle ABC stands on the diameter.

A proof must establish the result without assuming it, and use a valid congruence condition.

1. A radius is perpendicular to its tangent, so the triangles have right angles at A and B.
2. OA=OB because both are radii, and OP is the common hypotenuse.
3. Triangles OAP and OBP are congruent by RHS. Their corresponding sides give PA=PB.
4. Two equal sides alone do not give SSS; AAA gives similarity; the final option assumes the equality it is meant to prove.