Congruence and special triangles

Two triangles are congruent when they have the same shape and the same size — one can be laid on the other so that they coincide. The question that turns this into mathematics is: how much data has to be checked to be sure? The answer — three well-chosen items — is the first half of this lesson. The second half shows two triangles where a single number is enough, because the rest follows from fixed proportions.

The three congruence criteria

A triangle has six elements: three sides and three angles. It turns out that three of them — the right three — determine the rest.

SSS says that there is only one triangle with three given sides — which is why constructing a triangle from three segments always yields the same shape, and why a truss of triangles is rigid while a frame of quadrilaterals folds.

There is, however, no "side-side-angle" criterion, in which the angle does not lie between the given sides. Such data can fit two different triangles at once, so it settles nothing.

The notation $\triangle ABC \cong \triangle DEF$ carries more than the word "congruent": the order of the letters says which vertex corresponds to which, so it is immediately clear that $|AB| = |DE|$ and that the angle at $A$ equals the angle at $D$.

The 45°–45°–90° triangle is half a square

Cut a square of side $a$ along a diagonal. Two right triangles appear with two equal legs — and since such a triangle is isosceles, the base angles are equal and measure $45^\circ$ each.

The length of the diagonal comes from the Pythagorean theorem:

Hence the side ratio worth knowing by heart:

It works both ways. Given a leg, multiply by $\sqrt{2}$; given the hypotenuse, divide by $\sqrt{2}$, that is multiply by $\tfrac{\sqrt{2}}{2}$ — rationalising the denominator earns its keep in every exercise here.

The 30°–60°–90° triangle is half an equilateral one

This triangle is not half a square but half an equilateral triangle, and that explains all of its proportions at once.

Proof

Take an equilateral triangle of side $2a$ and drop the height from one vertex.

1. The height cuts the triangle into two triangles that are congruent by SSS: they share the height, each has one side $2a$, and each has half of the base.
2. Being congruent, they split the base into two equal parts, so the short leg is $a$, while the hypotenuse stays a side of the equilateral triangle, that is $2a$.
3. The base angle is $60^\circ$ (the triangle is equilateral), and the apex angle has been bisected, so it is $30^\circ$.
4. The other leg comes from the Pythagorean theorem:

Hence the whole relation, in which $a$ is the side opposite the $30^\circ$ angle:

The commonest mistake is attaching $a\sqrt{3}$ to the wrong side. The rule is simple: the larger angle faces the longer side, and $\sqrt{3} \approx 1.73$ sits between $1$ and $2$.

The height and area of an equilateral triangle

The proof above hands over the height of an equilateral triangle of side $a$ immediately: the short leg is $\tfrac{a}{2}$, so the height is $\tfrac{a}{2} \cdot \sqrt{3}$:

The area follows from the ordinary "base times height over two":

These are the only two formulas in this lesson worth memorising separately — everything else follows from the two side ratios.

Common mistakes

• Attaching $a\sqrt{3}$ to the short leg — the side opposite $30^\circ$ is $a$, and $a\sqrt{3}$ is opposite $60^\circ$.
• Confusing congruence with similarity — congruent triangles have equal sides, similar ones only proportional sides.
• Treating "side-side-angle" as a criterion — the angle must lie between the given sides.
• Dividing by $\sqrt{2}$ without rationalising — $\tfrac{12}{\sqrt{2}}$ is written as $6\sqrt{2}$.
• Taking half the side as the height of an equilateral triangle — the height is $\tfrac{a\sqrt{3}}{2}$, about $0.87a$, not $0.5a$.
• Computing that area as $\tfrac{a^2}{2}$ — the factor $\sqrt{3}$ is missing and the denominator is wrong.