Lesson · Congruence from the figure
Seeing congruence
Most municipal-round geometry ends with one sentence: these two triangles are congruent. The criteria are not the hard part - you know them. The work is deciding which two triangles to compare, and that decision is almost always made for you by the thing you are asked to prove.
The criteria, and the one that is not
| Name | What you must match |
|---|---|
| SSS | three sides |
| SAS | two sides and the angle between them |
| ASA / AAS | one side and two angles |
| RHS | right angle, hypotenuse, one leg |
Two sides and a non-included angle prove nothing. If your matching parts are \(AB = A'B'\), \(BC = B'C'\) and \(\angle BAC = \angle B'A'C'\), you have not finished - two different triangles fit that data.
Reading the pair off the conclusion
The conclusion names the objects you must compare, so let it choose the triangles:
- to prove two segments equal, find a triangle containing each and match them as corresponding sides;
- to prove two angles equal, do the same with the angles;
- to prove a right angle, look for an angle you can transport to the place where a right angle already sits.
Then hunt for the three matching parts. They come from the hypotheses, from shared sides, and from the standard suppliers of equal parts: a square or a regular polygon (all sides equal, all angles equal), a midpoint, a perpendicular bisector, an angle bisector, two radii of one circle.
Principle
A congruence is usually a symmetry of the configuration in disguise. Ask which rotation or reflection carries the first triangle onto the second - a quarter turn about the centre of a square, a reflection in a bisector, a half turn about a midpoint. Once you see the motion, the three matching parts write themselves.
Example
Let \(ABCD\) be a square, and let \(E\) on \(BC\) and \(F\) on \(CD\) satisfy \(BE = CF\). Prove that \(AE \perp BF\).
Nothing in the figure is a right angle yet except the corners of the square, so the \(90^\circ\) must be produced by an angle sum. Compare triangles \(ABE\) and \(BCF\): they have \(AB = BC\) (sides of the square), \(\angle ABE = \angle BCF = 90^\circ\) and \(BE = CF\) (hypothesis), so they are congruent by SAS. Hence
Now work inside triangle \(ABP\), where \(P = AE \cap BF\):
so \(\angle APB = 90^\circ\). The congruence is the quarter turn about the centre of the square, which sends \(A \mapsto B\), \(B \mapsto C\) and therefore the segment \(AE\) onto \(BF\) - a quarter turn always turns a line by \(90^\circ\).
The backwards trick
Some conclusions are not of the form "equal" at all: prove that these three points are collinear, or prove that this point lies on that line. A point is hard to argue about while it is defined as the intersection of two lines you cannot control. Redefine it.
Principle
To prove that a point \(F\) defined by one construction lies on a line \(\ell\), introduce \(F'\) as the point where \(\ell\) meets one of the lines defining \(F\), prove that \(F'\) has the defining property of \(F\), and conclude \(F = F'\) by uniqueness. You have replaced an unknown point by a point you built yourself.
Warning
The step \(F = F'\) needs uniqueness: two distinct lines meet in at most one point. Say which two lines you are intersecting, and check they really are distinct.
Warm-up
Example
On the sides \(AB\) and \(AC\) of a triangle \(ABC\), equilateral triangles \(ABX\) and \(ACY\) are drawn outwards. Prove that \(BY = CX\).
Which triangles even contain those two segments?
\(CX\) is a side of triangle \(XAC\), and \(BY\) is a side of triangle \(BAY\). Both have their third vertex at \(A\).
Look for the rotation
The rotation about \(A\) through \(60^\circ\) that sends \(X\) to \(B\) also sends \(C\) to \(Y\). That is the congruence; SAS is how you write it down.
The three matching parts
\(AX = AB\) and \(AC = AY\) because the small triangles are equilateral, and
SAS gives \(\triangle XAC \cong \triangle BAY\), hence \(XC = BY\).