Teach Me Maths

Congruent Triangles and Geometric Proof

Question 11 mark

Which statement describes two congruent triangles?

Hint

Consider which measurements must match after one triangle is moved.

Worked solution

Which statement describes two congruent triangles?

  1. They have equal areas but may have different shapes.
  2. They have the same size and shape.
  3. Their corresponding sides have any constant scale factor.
  4. They have the same perimeter but may have different angles.
  1. Congruent triangles have exactly the same size and shape. Corresponding sides and angles are equal. A rotation or reflection does not change this.

Question 21 mark

One triangle has side lengths 4, 6 and 8 cm. A second has side lengths 8, 4 and 6 cm. Which congruence criterion applies directly?

Hint

Match the three side lengths without relying on their order.

Worked solution

One triangle has side lengths 4, 6 and 8 cm. A second has side lengths 8, 4 and 6 cm. Which congruence criterion applies directly?

  1. ASA
  2. SAS
  3. RHS
  4. SSS
  1. Each of the three side lengths in one triangle has an equal partner in the other.
  2. The triangles are congruent by SSS.

Question 31 mark

In the triangles shown, AB=DE, AC=DF and \angle BAC=\angle EDF. Which congruence criterion applies directly?

Hint

Locate the equal angle relative to the two equal sides.

Worked solution

In the triangles shown, AB=DE, AC=DF and \angle BAC=\angle EDF. Which congruence criterion applies directly?

  1. SAS
  2. SSS
  3. RHS
  4. AAA
  1. The equal angles lie between the two pairs of equal sides.
  2. This is the side-angle-side criterion, SAS.

Question 41 mark

Triangles ABC and DEF have \angle A=\angle D=48^\circ, \angle B=\angle E=71^\circ and AB=DE=9 cm. Which criterion proves congruence directly?

Hint

Identify the side lying between the two stated angles.

Worked solution

Triangles ABC and DEF have \angle A=\angle D=48^\circ, \angle B=\angle E=71^\circ and AB=DE=9 cm. Which criterion proves congruence directly?

  1. RHS
  2. SSS
  3. ASA
  4. AAA
  1. Two pairs of corresponding angles are equal, and the side between them is equal.
  2. The triangles are congruent by ASA.

Question 51 mark

ABC and DEF are right-angled at B and E. Their hypotenuses satisfy AC=DF=13 cm, and AB=DE=5 cm. Which criterion applies directly?

Hint

Identify the hypotenuse and the additional equal side.

Worked solution

ABC and DEF are right-angled at B and E. Their hypotenuses satisfy AC=DF=13 cm, and AB=DE=5 cm. Which criterion applies directly?

  1. AAA
  2. RHS
  3. ASA
  4. SAS
  1. Both triangles are right-angled. Their hypotenuses are equal, and one other corresponding side is equal.
  2. This gives congruence by RHS.

Question 61 mark

Two triangles each have angles of 35^\circ, 65^\circ and 80^\circ. Which conclusion follows from this information alone?

Hint

Think about an enlargement of a triangle.

Worked solution

Two triangles each have angles of 35^\circ, 65^\circ and 80^\circ. Which conclusion follows from this information alone?

  1. They must be congruent by AAA.
  2. They must have the same area.
  3. They must have the same perimeter.
  4. They are similar, but may have different sizes.
  1. Equal corresponding angles establish similarity.
  2. An enlargement preserves angles but changes side lengths, so congruence is not guaranteed.

Question 71 mark

Triangle ABC is congruent to triangle PQR, with vertices corresponding in the stated order. Which side corresponds to AC?

Hint

Match both endpoints using the order of the triangle names.

Worked solution

Triangle ABC is congruent to triangle PQR, with vertices corresponding in the stated order. Which side corresponds to AC?

  1. PQ
  2. QR
  3. PR
  4. No side can be identified.
  1. A corresponds to P, and C corresponds to R.
  2. Therefore side AC corresponds to PR.

Question 82 marks

Triangle ABC is congruent to triangle PQR, in that order. BC=3x+2 cm and QR=20 cm. Find x.

Hint

First identify the two corresponding sides.

Worked solution

Triangle ABC is congruent to triangle PQR, in that order. BC=3x+2 cm and QR=20 cm. Find x.

  1. BC and QR correspond, so they are equal.
  2. 3x+2=20, giving 3x=18 and x=6.

Question 91 mark

A student claims that two equal sides and one equal non-included angle always prove two triangles congruent. Which response is correct?

Hint

Compare the claim with the requirements of SAS.

Worked solution

A student claims that two equal sides and one equal non-included angle always prove two triangles congruent. Which response is correct?

  1. The claim is false: the non-included angle does not always fix a unique triangle.
  2. The claim is true by SAS in every case.
  3. The claim is true because all triangles have three sides.
  4. The claim is false because equal angles can never help prove congruence.
  1. SAS needs the equal angle to be the one between the two equal sides.
  2. With a non-included angle, the third side can sometimes swing into two different positions. For example, with AB=10 cm, \angle A=40^\circ and BC=7 cm, point C can be in two places on the other arm of the angle.
  3. So two sides and a non-included angle do not always fix one triangle: the claim is false.

Question 101 mark

AB=DE and BC=EF. Which additional equality would allow a direct SAS proof that triangle ABC is congruent to triangle DEF?

Hint

The required angle must lie between the named sides in each triangle.

Worked solution

AB=DE and BC=EF. Which additional equality would allow a direct SAS proof that triangle ABC is congruent to triangle DEF?

  1. \angle BAC=\angle EDF
  2. \angle ACB=\angle DFE
  3. \angle BAC=\angle DFE
  4. \angle ABC=\angle DEF
  1. The sides AB and BC meet at B; DE and EF meet at E.
  2. The required equality is \angle ABC=\angle DEF.

Question 112 marks

AB=AC and M is the midpoint of BC. Which is a valid proof that triangles ABM and ACM are congruent?

Hint

Look for a side belonging to both triangles.

Worked solution

AB=AC and M is the midpoint of BC. Which is a valid proof that triangles ABM and ACM are congruent?

  1. AB=AC, BM=CM and AM is common, so SSS.
  2. The triangles share AM, so that single side is enough.
  3. Both triangles look the same, so no measurements are needed.
  4. The triangles have the same height, so they are congruent.
  1. AB=AC is given. BM=CM because M is the midpoint. AM is common.
  2. All three pairs of corresponding sides are equal, so the triangles are congruent by SSS.

Question 123 marks

In quadrilateral PQRS, PQ=PS and RQ=RS. The diagonal PR is drawn. \angle QPR=34^\circ. Find \angle QPS, in degrees.

Hint

Compare the triangles on either side of the diagonal.

Worked solution

In quadrilateral PQRS, PQ=PS and RQ=RS. The diagonal PR is drawn. \angle QPR=34^\circ. Find \angle QPS, in degrees.

  1. PQ=PS, RQ=RS, and PR is common, so triangles PQR and PSR are congruent by SSS.
  2. Corresponding angles give \angle RPS=\angle QPR=34^\circ.
  3. Therefore \angle QPS=34^\circ+34^\circ=68^\circ.

Question 133 marks

ABCD is a parallelogram. E is the midpoint of the diagonal AC. A straight line through E meets AB at P and DC at Q. Which is a complete proof that EP=EQ?

Hint

In triangles AEP and CEQ, look for one pair of equal sides and two pairs of equal angles, each with a reason.

Worked solution

ABCD is a parallelogram. E is the midpoint of the diagonal AC. A straight line through E meets AB at P and DC at Q. Which is a complete proof that EP=EQ?

  1. AE=CE and EP=EQ, so triangles AEP and CEQ are congruent (SAS). Hence EP=EQ.
  2. \angle PAE=\angle QCE and \angle APE=\angle CQE (alternate angles), so triangles AEP and CEQ are congruent (AAA). Hence EP=EQ.
  3. AE=CE, \angle PAE=\angle QCE (alternate angles) and \angle AEP=\angle CEQ (vertically opposite angles), so triangles AEP and CEQ are congruent (ASA). Hence EP=EQ.
  4. AE=CE, \angle PAE=\angle QCE (corresponding angles) and AP=CQ (opposite sides of a parallelogram), so triangles AEP and CEQ are congruent (SAS). Hence EP=EQ.
  1. Compare triangles AEP and CEQ.
  2. AE=CE because E is the midpoint of AC.
  3. AB is parallel to DC and AC crosses both, so \angle PAE=\angle QCE (alternate angles).
  4. PQ and AC cross at E, so \angle AEP=\angle CEQ (vertically opposite angles).
  5. The equal sides AE and CE lie between the two pairs of equal angles, so the triangles are congruent by ASA, and the corresponding sides EP and EQ are equal.
  6. The other proofs fail: one uses EP=EQ, the fact being proved; AAA only shows the triangles are similar; and AP=CQ is not known (only AB=DC is), and those angles are alternate, not corresponding.

Question 142 marks

AB=AC and AD is perpendicular to BC. A proof compares triangles ABD and ACD. After identifying the right angles and equal hypotenuses, which fact completes an RHS argument?

Hint

Identify a remaining side present in both triangles.

Worked solution

AB=AC and AD is perpendicular to BC. A proof compares triangles ABD and ACD. After identifying the right angles and equal hypotenuses, which fact completes an RHS argument?

  1. BD=DC because the diagram looks symmetrical.
  2. AB and AC must be parallel.
  3. The right angles alone prove congruence.
  4. AD is the same side in both triangles.
  1. AB and AC are equal hypotenuses. Both triangles have a right angle at D.
  2. AD is common, providing the additional equal side for RHS.

Question 152 marks

ABCD is a rectangle. Triangles ABC and BAD have AB in common, BC=AD and equal included right angles. Which conclusion follows by SAS?

Hint

Match the remaining sides after identifying the correspondence.

Worked solution

ABCD is a rectangle. Triangles ABC and BAD have AB in common, BC=AD and equal included right angles. Which conclusion follows by SAS?

  1. AC=AB
  2. AC=BD
  3. AC=BC
  4. AB=BC
  1. AB is common, BC=AD, and the included angles ABC and BAD are right angles.
  2. The triangles are congruent by SAS. Their corresponding third sides give AC=BD.

Question 16Challenge3 marks

AB and CD intersect at M, the midpoint of both segments. AC=4x+1 cm and BD=7x-8 cm. Find x.

Hint

Compare the two triangles using the angles at the intersection.

Worked solution

AB and CD intersect at M, the midpoint of both segments. AC=4x+1 cm and BD=7x-8 cm. Find x.

  1. AM=BM and CM=DM because M is a midpoint.
  2. \angle AMC=\angle BMD because vertically opposite angles are equal. Thus triangles AMC and BMD are congruent by SAS.
  3. AC=BD, so 4x+1=7x-8. Therefore 3x=9 and x=3.

Question 17Challenge3 marks

Equilateral triangles ABC and ABD lie on opposite sides of their common side AB. The segment CD is drawn. Find \angle ACD, in degrees.

Hint

Compare triangles ACD and BCD using their sides.

Worked solution

Equilateral triangles ABC and ABD lie on opposite sides of their common side AB. The segment CD is drawn. Find \angle ACD, in degrees.

  1. AC=BC and AD=BD because the two triangles are equilateral. CD is common.
  2. Triangles ACD and BCD are congruent by SSS, so the angles at C are equal.
  3. \angle ACB=60^\circ, so \angle ACD=60^\circ\div2=30^\circ.

Question 18Challenge4 marks

In triangle ABC, AB=AC and AD is perpendicular to BC. AB=5x-4 cm, AC=3x+8 cm and BD=2x+1 cm. Find BC, in cm.

Hint

Use the equal sides to find x, then justify how the perpendicular splits the base.

Worked solution

In triangle ABC, AB=AC and AD is perpendicular to BC. AB=5x-4 cm, AC=3x+8 cm and BD=2x+1 cm. Find BC, in cm.

  1. 5x-4=3x+8 gives 2x=12, so x=6.
  2. BD=2(6)+1=13 cm.
  3. Triangles ABD and ACD are congruent by RHS: equal hypotenuses AB=AC, common side AD and right angles at D.
  4. Hence DC=BD=13 cm and BC=26 cm.

Question 19Challenge4 marks

ABCD is an isosceles trapezium with AB\parallel DC, AB=12 cm and AD=BC=13 cm. Perpendiculars AX and BY meet DC at X and Y; AX=BY=5 cm. The points on the longer base are in the order D,X,Y,C. Find DC, in cm.

Hint

Compare the right-angled triangles at the ends and use the rectangle in the middle.

Worked solution

ABCD is an isosceles trapezium with AB\parallel DC, AB=12 cm and AD=BC=13 cm. Perpendiculars AX and BY meet DC at X and Y; AX=BY=5 cm. The points on the longer base are in the order D,X,Y,C. Find DC, in cm.

  1. AXYB is a rectangle, so XY=AB=12 cm.
  2. Triangles ADX and BCY have equal hypotenuses, equal perpendicular heights and right angles. They are congruent by RHS, so DX=YC.
  3. Pythagoras gives DX=\sqrt{13^2-5^2}=\sqrt{144}=12 cm.
  4. DC=DX+XY+YC=12+12+12=36 cm.

Question 20Challenge3 marks

AB=AC. Points D and E lie on AB and AC respectively, with AD=AE. Which complete argument establishes BE=CD?

Hint

Choose two triangles with the angle at A in common.

Worked solution

AB=AC. Points D and E lie on AB and AC respectively, with AD=AE. Which complete argument establishes BE=CD?

  1. Triangles ABE and ACD have two equal sides, which alone prove congruence.
  2. AB=AC, AE=AD and \angle BAE=\angle CAD give SAS for triangles ABE and ACD, hence BE=CD.
  3. Triangles BDE and CDE share DE, so they must be congruent.
  4. \angle BAE=\angle CAD gives AAA, which proves the triangles are congruent.
  1. Compare triangles ABE and ACD. AB=AC and AE=AD are given.
  2. The included angles \angle BAE and \angle CAD are the same angle at A, since D lies on AB and E lies on AC.
  3. The triangles are congruent by SAS, so the corresponding sides BE and CD are equal.

More on this topic: Congruent Triangles and Geometric Proof worksheet with full solutions

All Edexcel GCSE Higher practice questions