G5, G6 - Congruent triangles and geometric proof

G5, G6 - Congruent triangles and geometric proof

Decide when triangle measurements force exactly one shape. G5 supplies four congruence tests; G6 uses them, angle facts, quadrilateral properties and similarity to build proofs. You will finish by explaining why the squares on the two shorter sides of a right-angled triangle add to the square on its longest side.

Congruence from sides and included angles

Four identical right-angled triangles fit inside a square of side 77. Each has shorter sides 33 and 44. What is the area of the tilted square left in the middle? By the end, this little packing puzzle will explain a famous theorem.

You already know

Angles in a triangle add to 180∘180^\circ. An isosceles triangle has two equal sides; a parallelogram has two pairs of parallel opposite sides. Corresponding points on a coordinate grid keep their labels when a shape moves.

First we need to recognise identical triangles, even when one is turned over. Matching the longest side to the longest side is a useful start, but the positions of the angles matter too.

Congruent triangles

Triangles with exactly the same shape and size. Corresponding sides and corresponding angles are equal. A rotation, reflection or translation can place one exactly on the other. A larger copy is similar, but is not congruent.

You do not need to measure all six features. A small set of matching measurements can force the rest.

SSS and SAS congruence

SSS: three pairs of corresponding sides are equal.

SAS: two pairs of corresponding sides and the included angle, between those sides, are equal.

These tests apply to triangles with positive area. For SSS, the longest side must be shorter than the sum of the other two.

Not on the formulae sheet: learn it.

With SSS, once one side is fixed, the third vertex is where two circles meet. The two possible positions are mirror images, so they still give congruent triangles. With SAS, the included angle fixes the opening before the two side lengths are marked.

[DIAGRAM: asset_name: Three sides or two sides and their included angle; asset_slug: aqa8300_g05_sss_sas; caption: SSS fixes three side lengths and SAS fixes the angle between the two known sides; description: Two panels. The first has a pair of triangles with corresponding sides 4 5 6 in different orientations. The second has a triangle with sides 5 and 7 and the included angle 60 degrees marked. A note says the known sides meet at A. Diagrams not drawn accurately.]
SSS fixes three side lengths and SAS fixes the angle between the two known sides

Worked example 1: matching three sides

Question. Triangle ABCABC has AB=4AB=4, BC=6BC=6, CA=5CA=5. Triangle PQRPQR has PQ=6PQ=6, QR=5QR=5, RP=4RP=4. Explain why the triangles are congruent and write their vertices in corresponding order. Non-calculator.

Plan. Match all three sides, then use the sides meeting at each vertex.

AB=RP=4first side pairBC=PQ=6second side pairCA=QR=5third side pair\begin{aligned} AB&=RP=4 &&\text{first side pair}\\ BC&=PQ=6 &&\text{second side pair}\\ CA&=QR=5 &&\text{third side pair} \end{aligned}

Why this step? Vertex AA joins the sides of lengths 44 and 55, so it matches RR, not PP.

Answer. △ABC\triangle ABC is congruent to △RPQ\triangle RPQ, by SSS.

Check. The order gives A↔RA\leftrightarrow R, B↔PB\leftrightarrow P, C↔QC\leftrightarrow Q, which matches every side.

Change only the evidence: instead of the third side, the next example gives the angle between the first two sides.

Worked example 2: using the included angle

Question. AB=DE=5AB=DE=5, AC=DF=7AC=DF=7, and ∠BAC=∠EDF=60∘\angle BAC=\angle EDF=60^\circ. Explain why △ABC\triangle ABC and △DEF\triangle DEF are congruent. Non-calculator.

Plan. Check that the equal angles lie between the two equal side pairs.

AB=DEgiven sides∠BAC=∠EDFgiven included anglesAC=DFgiven sides\begin{aligned} AB&=DE &&\text{given sides}\\ \angle BAC&=\angle EDF &&\text{given included angles}\\ AC&=DF &&\text{given sides} \end{aligned}

Why this step? The middle letters AA and DD are where the two known sides meet. This is SAS, rather than two sides and an unrelated angle.

Answer. △ABC\triangle ABC is congruent to △DEF\triangle DEF, by SAS.

Check. Both angle statements use the vertex between the sides of lengths 55 and 77.

Two sides and a different angle may allow two different triangles. This is why the position of an angle must be stated precisely.

Watch out

Wrong: “Two sides and any equal angle give SAS.” Right: the equal angle must be between the two equal sides. SSA is not a general congruence test.

Use the named vertices to check the included angle in this new pair.

Congruence from angles and right triangles

Two angles fix a triangle's shape, but not its size. Adding a corresponding side fixes the size as well.

ASA and RHS congruence

ASA: two corresponding angles and the included side are equal. If the known side is not included, first find the third angle using 180∘180^\circ; this gives an ASA arrangement.

RHS: both triangles are right-angled, their hypotenuses are equal, and one other corresponding side is equal. The hypotenuse is the side opposite the right angle.

Not on the formulae sheet: learn it.

AAA is a useful test for similarity, but not for congruence: an equilateral triangle can have side 22 or side 2020. RHS is a special right-triangle test; it does not make SSA valid for other triangles.

[DIAGRAM: asset_name: Angle side angle and right angle hypotenuse side; asset_slug: aqa8300_g05_asa_rhs; caption: ASA uses corresponding angles and a side while RHS requires two right angles and equal hypotenuses; description: Two panels. The first shows a triangle with angles 42 and 68 degrees at the ends of a side of length 9. The second shows two right-angled triangles with hypotenuse 13 and one shorter side 5, in different orientations. Right-angle squares and hypotenuse labels are clear. Diagrams not drawn accurately.]
ASA uses corresponding angles and a side while RHS requires two right angles and equal hypotenuses

Worked example 3: finding the angle needed for ASA

Question. In ABCABC, ∠A=42∘\angle A=42^\circ, ∠B=68∘\angle B=68^\circ, AB=9AB=9. In PQRPQR, ∠P=42∘\angle P=42^\circ, ∠R=70∘\angle R=70^\circ, PQ=9PQ=9. Explain why the triangles are congruent. Non-calculator.

Plan. Find ∠Q\angle Q, then compare the angles at the ends of the equal sides.

∠Q=180∘−42∘−70∘triangle angle sum=68∘so it matches angle B∠A=∠P,AB=PQ,∠B=∠QASA data\begin{aligned} \angle Q&=180^\circ-42^\circ-70^\circ &&\text{triangle angle sum}\\ &=68^\circ &&\text{so it matches angle }B\\ \angle A&=\angle P,\quad AB=PQ,\quad\angle B=\angle Q &&\text{ASA data} \end{aligned}

Why this step? The new angle puts the equal side between the two equal angles in each triangle.

Answer. △ABC\triangle ABC is congruent to △PQR\triangle PQR, by ASA.

Check. Each third angle is 70∘70^\circ; all three angle pairs match.

Change the triangle type to right-angled. The right angle and hypotenuse now let you use RHS directly.

Worked example 4: recognising the hypotenuse

Question. ABCABC is right-angled at BB, with AC=13AC=13 and AB=5AB=5. DEFDEF is right-angled at EE, with DF=13DF=13 and DE=5DE=5. Explain why the triangles are congruent. Non-calculator.

Plan. Identify the hypotenuses from the right-angle positions, then match the other given side.

∠B=∠E=90∘right anglesAC=DF=13hypotenusesAB=DE=5another corresponding side\begin{aligned} \angle B&=\angle E=90^\circ &&\text{right angles}\\ AC&=DF=13 &&\text{hypotenuses}\\ AB&=DE=5 &&\text{another corresponding side} \end{aligned}

Why this step? A hypotenuse is identified by being opposite the right angle, not by how a drawing is tilted.

Answer. △ABC\triangle ABC is congruent to △DEF\triangle DEF, by RHS.

Check. B↔EB\leftrightarrow E matches the right angles, and A↔DA\leftrightarrow D matches the ends of the sides of length 55.

In a proof, name the criterion after listing the matching information. “They look the same” is not mathematical evidence.

Watch out

Wrong: “The longest-looking sides are equal, so RHS.” Right: establish a right angle in each triangle, identify their hypotenuses and give one more equal side pair. A sketch alone establishes none of these facts.

The next pair has the same right-triangle structure, with new lengths and a different vertex order.

Proving angle and side properties

A useful proof often begins by drawing one extra line. It splits a familiar shape into triangles whose matching facts can be explained.

A conjecture is a statement you think may be true. A diagram can suggest one, but measuring several examples cannot prove it for every shape. Begin with the facts given, name a reason for each new fact, and reach the claim without using it along the way.

Geometric proof

A chain of deductions from given information and established results. For example, “these sides are equal because the triangles are congruent” is a valid step after congruence has been proved. “These triangles are congruent because the sides I want to prove equal are equal” is circular.

We will derive a triangle property, then use the same strategy inside a quadrilateral. These proof methods are not on the formulae sheet: learn them.

[DIAGRAM: asset_name: Extra lines reveal matching triangles; asset_slug: aqa8300_g05_proof_triangles; caption: An angle bisector in an isosceles triangle and a diagonal in a parallelogram create congruent triangles; description: Left an isosceles triangle ABC with apex A, AB and AC marked equal, and AD bisecting the apex angle to a point D on BC. Right a general parallelogram PQRS with diagonal PR, parallel side arrow marks and corresponding alternate angles shaded. Diagrams not drawn accurately.]
An angle bisector in an isosceles triangle and a diagonal in a parallelogram create congruent triangles

Worked example 5: why isosceles base angles are equal

Question. AB=ACAB=AC. Point DD lies on BCBC, and ADAD bisects ∠BAC\angle BAC. Prove that ∠ABC=∠BCA\angle ABC=\angle BCA.

Plan. Prove the two triangles either side of ADAD congruent, then match their base angles.

AB=ACgiven∠BAD=∠DACangle bisectorAD=ADcommon side\begin{aligned} AB&=AC &&\text{given}\\ \angle BAD&=\angle DAC &&\text{angle bisector}\\ AD&=AD &&\text{common side} \end{aligned}

So △ABD\triangle ABD and △ACD\triangle ACD are congruent by SAS. Their corresponding angles at BB and CC are equal.

Why this step? The common side is genuine evidence. There is no need to assume that DD is a midpoint or that ADAD is perpendicular to BCBC.

Answer. ∠ABC=∠BCA\angle ABC=\angle BCA: the base angles of an isosceles triangle are equal.

Check. Congruence also gives BD=DCBD=DC and equal adjacent angles at DD. Those angles add to 180∘180^\circ, so each is 90∘90^\circ, consistent with the symmetry.

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