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Congruent triangles

Geometry · level 5, lesson 5 of 6

A question to start

You want to know how wide a river is, but you cannot cross it. On the far bank, right opposite you, there is a tree. All you have is a long tape measure, a few sticks and your feet. Can you find the width without getting wet?

Yes, and people have used tricks like this for thousands of years. You walk along your own bank, plant a stick, walk the same distance again, turn at a right angle away from the river, and walk until the stick and the tree line up. The distance you walked away from the river is the width of the river. Worked example 4 shows why. The reason is a pair of triangles that are exact copies of each other, one across the water and one on dry land.

The idea

What "congruent" means

Two shapes are congruent if one can be moved exactly on top of the other by translations, rotations and reflections (the rigid motions of lesson 3). Congruent shapes have the same size and the same shape; only their position, and maybe which way round they face, is different.

For triangles we write \(\triangle ABC \cong \triangle DEF\), read "triangle \(ABC\) is congruent to triangle \(DEF\)". The order of the letters matters: it says which corner matches which. \(A\) matches \(D\), \(B\) matches \(E\), \(C\) matches \(F\). So side \(AB\) matches \(DE\), angle \(C\) matches angle \(F\), and so on. Matching sides and angles are called corresponding parts.

If two triangles are congruent, all six corresponding parts are equal: three sides and three angles. The big question of this lesson is the other way round: how few equal parts do you need to check to be sure two triangles are congruent?

Three parts are enough, if they are the right three

In lesson 2 you saw that some sets of three measurements let you build only one triangle (maybe flipped over). That is exactly what we need: if two triangles share those three measurements, they were both built by the same recipe, so they must be copies of each other. Each recipe becomes a congruence test.

SSS (side, side, side). If the three sides of one triangle equal the three sides of the other, the triangles are congruent. Why: with the base drawn, the third corner must lie on two circles, which cross only at two mirror-image points. Any two triangles with those sides are copies.

SAS (side, angle, side). Two sides and the angle between them (the included angle). Why: the angle fixes the direction of the second side and its length fixes the third corner; there is no choice left.

ASA (angle, side, angle). Two angles and the side between them. Why: two rays from the ends of a fixed side, in fixed directions, cross at only one point.

AAS (angle, angle, side). Two angles and a side that is not between them. Why: the angles add to 180°, so the third angle is known too; now the side is between two known angles, and ASA applies.

RHS (right angle, hypotenuse, side). Both triangles have a right angle, equal hypotenuses (the side opposite the right angle, always the longest) and one other pair of equal sides. Why: draw the right angle and mark the known leg along one arm. The third corner is on the other arm, and it must be one hypotenuse-length from the top of the leg. A circle of that radius around the top of the leg crosses the other arm's line once on each side of the right angle, and those two points are mirror images. So the triangle is fixed. (In level 6 you will also be able to calculate the third side with Pythagoras.)

Why SSA fails

SSA means two sides and an angle that is not between them. Try \(\angle A = 30°\), \(AB = 10\) and \(BC = 6\).

Draw the angle at \(A\). Mark \(B\) on one arm, 10 from \(A\). The corner \(C\) is on the other arm, 6 from \(B\), so it is on the circle of radius 6 around \(B\). That circle crosses the arm twice:

ABC₁C₂ 30°10665

Both triangles \(ABC_1\) (blue) and \(ABC_2\) (blue and orange together) have \(\angle A = 30°\), \(AB = 10\) and \(BC = 6\). But they are clearly not the same: one is much wider than the other. So SSA is not a congruence test.

How high is \(B\) above the base? A right triangle with a 30° angle is exactly half of an equilateral triangle (cut an equilateral triangle down its line of symmetry), so the side opposite the 30° angle is half the hypotenuse. Here the height is \(10 \div 2 = 5\). Now compare \(BC\) with that height of 5 and with \(AB = 10\):

\(BC\) What the circle does Triangles
less than 5 does not reach the base none
exactly 5 just touches the base one (a right triangle)
between 5 and 10 crosses the base twice two
10 or more crosses the arm only once (the other crossing is at \(A\) or behind it) one

So SSA does work in one situation: when the side opposite the given angle is at least as long as the other given side. RHS is that case, since the hypotenuse is always the longest side. This "two triangles" puzzle returns in level 7 as the ambiguous case of the sine law.

AAA is not a test either: three equal angles give the same shape but any size.

Using congruence to prove things

Once two triangles are known to be congruent, every pair of corresponding parts is equal, including the ones you were not given. This makes congruence a machine for proving facts. Three classic examples:

The base angles of an isosceles triangle are equal. An isosceles triangle has two equal sides, \(AB = AC\). Draw the bisector of angle \(A\), meeting \(BC\) at \(M\).

ABCM

In triangles \(ABM\) and \(ACM\): \(AB = AC\) (given), \(\angle BAM = \angle CAM\) (we bisected), and \(AM\) is shared. That is SAS, so \(\triangle ABM \cong \triangle ACM\), and therefore \(\angle B = \angle C\). As a bonus, \(BM = CM\) and the two angles at \(M\) are equal and add to 180°, so each is 90°: the bisector from the top is also the perpendicular bisector of the base.

Every point on the perpendicular bisector of \(AB\) is equally far from \(A\) and \(B\). Take \(P\) on it, with \(M\) the midpoint. Triangles \(PMA\) and \(PMB\) have \(AM = BM\), right angles at \(M\), and \(PM\) shared: SAS. So \(PA = PB\). This is the fact that made the constructions of lesson 1 work.

Opposite sides of a parallelogram are equal. Draw a diagonal. It makes two triangles with the diagonal shared, and two pairs of alternate angles equal (from level 4: a line crossing two parallel lines makes equal alternate angles). That is ASA, so the triangles are congruent and the opposite sides match.

Worked examples

1. Which test? In each case, say whether the triangles must be congruent.

  1. \(AB = DE = 5\), \(BC = EF = 7\), \(\angle B = \angle E = 40°\). The angle is between the two sides: SAS, congruent.
  2. \(\angle A = \angle D = 50°\), \(\angle B = \angle E = 60°\), \(BC = EF = 8\). Two angles and a side not between them: AAS, congruent.
  3. \(AB = DE = 9\), \(BC = EF = 6\), \(\angle A = \angle D = 30°\). The angle is not between the sides: SSA, not enough. By the table, \(6\) is between half of 9 and 9, so there are two possible triangles.

2. Using the matching. \(\triangle PQR \cong \triangle XYZ\), with \(PQ = 7\) cm, \(QR = 9\) cm and \(\angle R = 40°\). From the order of letters, \(Q \leftrightarrow Y\) and \(R \leftrightarrow Z\), so \(YZ = QR = 9\) cm and \(\angle Z = \angle R = 40°\).

3. Diagonals of a rectangle are equal. Rectangle \(ABCD\): compare triangles \(ABC\) and \(BAD\). \(AB\) is shared; \(BC = AD\) (opposite sides of a rectangle); \(\angle ABC = \angle BAD = 90°\). SAS, so the triangles are congruent and the diagonals \(AC\) and \(BD\) are equal.

4. The river. You stand at \(A\), right opposite a tree \(T\), so \(AT\) crosses the river at right angles. Walk along the bank 40 m to \(B\) and plant a stick. Walk on 40 m to \(C\). Turn 90° away from the river and walk until the stick at \(B\) hides the tree. Call that spot \(D\). Suppose \(CD = 23\) m.

  1. \(AB = CB = 40\) m.
  2. \(\angle TAB = \angle DCB = 90°\).
  3. \(\angle ABT = \angle CBD\): they are vertically opposite angles, because \(T\), \(B\), \(D\) are in a line.
  4. ASA: \(\triangle ABT \cong \triangle CBD\). So \(AT = CD = 23\) m. The river is 23 m wide.

Try it

  1. Name the test, or say "not enough": (a) all three sides equal; (b) two angles and a side not between them; (c) \(AB = DE\), \(BC = EF\), \(\angle A = \angle D\); (d) all three angles equal; (e) both have a right angle, equal hypotenuses and one other equal side.
  2. \(\triangle KLM \cong \triangle RST\). \(KL = 4\), \(LM = 6\), \(\angle K = 75°\), \(\angle L = 45°\). Find \(ST\), \(\angle R\) and \(\angle T\).
  3. A kite \(ABCD\) has \(AB = AD\) and \(CB = CD\). Prove that the diagonal \(AC\) bisects \(\angle A\).
  4. \(\angle A = 30°\) and \(AB = 8\). For each length of \(BC\), how many triangles \(ABC\) are there? (a) 3 (b) 4 (c) 6 (d) 10.
  5. In the river method, \(AB = CB = 25\) m and \(CD = 18\) m. How wide is the river?
  6. Why are any two equilateral triangles with sides of 5 cm congruent?
  7. Harder: \(P\) is on the bisector of angle \(A\). Drop perpendiculars from \(P\) to the two arms, meeting them at \(X\) and \(Y\). Prove that \(PX = PY\).
Answers
  1. (a) SSS. (b) AAS. (c) Not enough: SSA. (d) Not enough: AAA fixes the shape, not the size. (e) RHS.
  2. \(K \leftrightarrow R\), \(L \leftrightarrow S\), \(M \leftrightarrow T\). \(ST = LM = 6\), \(\angle R = \angle K = 75°\), and \(\angle T = \angle M = 180° - 75° - 45° = 60°\).
  3. In triangles \(ABC\) and \(ADC\): \(AB = AD\), \(CB = CD\), and \(AC\) is shared. SSS, so they are congruent, and \(\angle BAC = \angle DAC\). So \(AC\) cuts \(\angle A\) into two equal parts.
  4. The height of \(B\) above the other arm is \(8 \div 2 = 4\). (a) 3 is less than 4: no triangle. (b) 4: exactly one, a right triangle. (c) 6 is between 4 and 8: two triangles. (d) 10 is at least 8: one triangle.
  5. 18 m, by the same ASA argument as worked example 4.
  6. All three sides are 5 cm in both: SSS.
  7. In triangles \(AXP\) and \(AYP\): \(\angle XAP = \angle YAP\) (\(AP\) is the bisector), \(\angle AXP = \angle AYP = 90°\), and \(AP\) is shared (it is not between the two angles in either triangle). AAS, so the triangles are congruent, and \(PX = PY\). Every point on an angle bisector is equally far from the two arms.

Common mistake

Writing the letters in the wrong order. If \(\angle A\) matches \(\angle E\), do not write \(\triangle ABC \cong \triangle DEF\). The statement promises \(A \leftrightarrow D\), so every conclusion you draw from it would pair the wrong parts. Match corners first, then write the letters in matching order.

Using SSA or AAA as a test. Both are tempting because they are three equal parts. Neither forces the triangles to be the same. Check: is the angle between the two sides (SAS)? If not, and there is no right angle with a hypotenuse (RHS), you cannot conclude congruence.

Where it leads