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Congruence, similarity and circle proofs

Choose the triangles that contain the target, justify congruence or similarity with exact correspondences and use circle properties only in their stated configurations.

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Enlarged diagram

A geometric proof explains why something must be true for every figure of a given kind, not just the one drawn. Congruent and similar triangles are the usual tools, often combined with circle properties. This lesson shows how to choose the triangles, state each condition, and write a proof another reader can check.

Choose the triangles that contain the target

To prove two lengths or two angles equal, find two triangles that each contain one of them. Then show that the triangles are congruent. Write the vertices in corresponding order, so the statement itself says which parts match.

A congruence needs one of these tests:

  • SSS: three pairs of equal sides.
  • SAS: two pairs of equal sides and the angles between them.
  • ASA or AAS: two pairs of equal angles and one pair of corresponding sides.
  • RHS: a right angle, equal hypotenuses and one other pair of equal sides.

Two sides and an angle that is not between them is not a test. It can fit two different triangles.

Worked example 1

Prove a second pair of sides parallel

In quadrilateral ABCD, AB = CD and AB∥ DC. Draw diagonal BD and prove that AD∥ BC.

Only AB and DC are marked. Whether AD is parallel to BC is what the proof decides.
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  1. Build the congruence from the givens

    Method

    Compare triangle ABD and triangle CDB.

    Reason

    The diagonal supplies a common side, and the given parallel pair supplies the included equal angles.

    Working

    Answer

    We have AB = CD (given), ∠ ABD = ∠ CDB (alternate angles since AB∥ DC), and BD = DB (common side). SAS therefore gives triangle ABD≡triangle CDB, with A ↔ C, B ↔ D and D ↔ B.

  2. Use the corresponding angles

    Method

    Read ∠ ADB = ∠ CBD from the congruence.

    Reason

    Congruent triangles have equal corresponding angles; these angles lie across transversal BD.

    Working

    Answer

    The equal alternate angles imply AD∥ BC by the converse of the alternate-angle property.

  3. State the result

    Working

    Answer

    Thus AD∥ BC. Both pairs of opposite sides are parallel, so ABCD is a parallelogram.

Check your understanding 1

The vertex order is part of the claim

For the same quadrilateral, a student writes triangle ABD≡triangle BCD. Explain what is wrong with this statement and give one false equality it would claim.

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The reasons pair A with C, B with D and D with B. The statement triangle ABD≡triangle BCD instead pairs A with B, B with C and D with D. It would claim, for example, that AB = BC and ∠ BAD = ∠ CBD. Neither follows from the givens. The correct statement is triangle ABD≡triangle CDB.

Tangents from one point give congruent right-angled triangles

The right angles come from the tangent–radius property. The equal lengths are proved, not read from the drawing.

PA and PB are tangents from P to the circle with centre O.

  1. OA = OB (radii).
  2. ∠ OAP = ∠ OBP = 90° (a tangent is perpendicular to the radius at the point of contact).
  3. OP is the hypotenuse of both triangles (common side).
  4. So triangle OAP≡triangle OBP (RHS), with O ↔ O, A ↔ B, P ↔ P.

Hence PA = PB, and ∠ APO = ∠ BPO, so OP bisects ∠ APB. The same congruence also gives ∠ AOP = ∠ BOP.

Check your understanding 2

Use the right angles in the quadrilateral

In the figure, ∠ APB = 50°. Find ∠ AOB. Then explain why quadrilateral OAPB has two pairs of equal adjacent sides.

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The angles of quadrilateral OAPB add to 360°, and the angles at A and B are both 90°. So

∠ AOB = 360°-90°-90°-50°; = 130°.

OA = OB because they are radii. PA = PB from the congruent triangles above. So OAPB is a kite, with OP as its line of symmetry.

Use a circle property only in its stated configuration

Each circle property has a condition. Check it before you quote the property.

  • Angle in a semicircle: the angle is 90° only when its arms end at the two ends of a diameter.
  • Angle at the centre: it is twice an angle at the circumference only when both stand on the same arc.
  • Angles in the same segment: they are equal only when both vertices are on the same side of the chord.
  • Cyclic quadrilateral: opposite angles add to 180° only when all four vertices lie on the circle.
  • Tangent and radius: they are perpendicular at the point of contact.

In a proof, name the chord or arc as well as the property. For example, write “∠ CAB = ∠ CDB (angles in the same segment, chord CB)”. If the two vertices are on opposite sides of the chord, the angles add to 180° instead.

Similar triangles turn angle facts into products

A and D are on the same side of chord CB. The diagram shows the arrangement; it does not give any lengths.

Chords AB and CD meet at X inside the circle. Compare triangles AXC and DXB.

  1. ∠ AXC = ∠ DXB (vertically opposite angles).
  2. ∠ CAX = ∠ BDX, because ∠ CAB = ∠ CDB (angles in the same segment, chord CB).
  3. So triangle AXC∼triangle DXB (two pairs of equal angles), with A ↔ D, X ↔ X, C ↔ B.

Corresponding sides are in the same ratio:

AX/(DX) = XC/(XB).

Multiplying both sides by DX × XB gives AX × XB = CX × XD. The two parts of one chord multiply to the same value as the two parts of the other.

Check your understanding 3

Match the corresponding sides

In the figure, AX = 5 cm, XB = 4 cm and CX = 2 cm. Find XD. Then explain why the equation 5/4 = 2/(XD) is not correct.

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From the similarity, AX/(DX) = XC/(XB), so AX × XB = CX × XD:

5 × 4 = 2 × XD,; XD = 10 cm.

The equation 5/4 = 2/(XD) compares AX with XB, which are two parts of the same chord. They are not corresponding sides of the similar triangles. That equation would give XD = (8/5) cm, which is wrong.

Try it yourself 1

Prove a length relation in a semicircle

Minimal support

AB is a diameter of a circle and C is another point on the circle. D is the point on AB where CD is perpendicular to AB. Prove that CD² = AD × DB, giving a reason for each step. Then decide whether this relation still holds when C is not on the circle, and justify your answer.

The right angle at C is a consequence of AB being a diameter. Use the stated facts rather than measurements.

Write your answer, then compare it with the solution

Hint 1

Look for two right-angled triangles that share an angle relationship

Triangles ADC and CDB each have a right angle at D. Use the right angle at C to connect one of their other angles.

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  1. Identify the right angles

    Method

    Find every right angle that the givens supply.

    Reason

    AB is a diameter and CD is perpendicular to AB.

    Working

    Answer

    ∠ ACB = 90° (angle in a semicircle, AB is a diameter). ∠ ADC = ∠ CDB = 90° (CD⊥ AB).

  2. Match a second pair of angles

    Method

    Show that ∠ DAC = ∠ DCB.

    Reason

    Both angles make 90° with the same angle ∠ ACD.

    Working

    Answer

    In triangle ADC the angle at D is 90°, so ∠ DAC + ∠ ACD = 90°. At C, ∠ ACD + ∠ DCB = ∠ ACB = 90°. Subtracting ∠ ACD from both equations gives

    ∠ DAC = ∠ DCB.

  3. State the similarity and use its ratio

    Method

    Write the similarity with its correspondence, then the ratio of corresponding sides.

    Reason

    Two pairs of equal angles prove that the triangles are similar.

    Working

    Answer

    ∠ ADC = ∠ CDB and ∠ DAC = ∠ DCB, so triangle ADC∼triangle CDB, with A ↔ C, D ↔ D, C ↔ B. Hence

    AD/(CD) = DC/(DB),; CD² = AD × DB.

  4. Test the condition that C is on the circle

    Method

    Keep the perpendicular but move C off the circle.

    Reason

    The proof used ∠ ACB = 90°, which comes from C being on the circle.

    Working

    Answer

    Take A(0,0), B(4,0), C(1,1). Then D is (1,0), so CD = 1 while AD × DB = 1 × 3 = 3. Here ∠ ACB is not 90°: the vectors from C to A and from C to B are (-1,-1) and (3,-1), whose scalar product is -2, not 0. So the relation fails without the circle condition.

Common mistakes

  • Naming triangles with unmatched vertices. In triangle PAO≡triangle PBO, each letter must match its partner.
  • Using SSA as a congruence test. Two sides and a non-included angle do not fix a triangle.
  • Quoting a circle property outside its condition. For example, the angle in a semicircle needs a diameter.
  • Assuming what you must prove. Start from the given facts.

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