Figure 0.
So here's the problem to be solved: Given an arbitrary figure, let's say a polygon, and an arbitrary line, find the reflection of that polygon around the arbitrary line. The problem is illustrated in Figure 0. ["Figure zero?" you ask. "Why zero?" Well, think about it. We're not to step 1; you haven't done anything yet.]
Specifically, we have a quadrilateral ABCD and a line r that doesn't intersect. And no coordinate system.
We could have a problem where r does intersect ABCD. Any point on line r would be its own reflection, and the rest of the solution would look just about the same.
Let's begin by reflecting the vertex A across the arbitrary line r. We have to start somewhere, so we might as well start with A.
We need a line through A that's perpendicular to r. (That's from definition of reflection.) We need to find the point A′ on that new line which is exactly the same distance from the line as A.
Fortunately, that's really easy to do. All we have to do is construct a line perpendicular to r through point A, which is something we've done before.
Draw a circle centered on point A and intersecting r in 2 points, J and K. This is shown in Figure 1. The circle can be any size you want, so long as the radius is bigger than the distance to the line r. (Otherwise you won't have J and K, and you need them.)
Figure 1.
If the circle were too small, it wouldn't reach to the line and there would be no intersection. If the radius were just exactly the distance to the line, the circle and line would intersect in only one point. (The line would be a tangent.)
We pick a bigger length for the radius so that we can be sure that it will intersect the line in 2 points.
Around each of the new points J and K, draw another circle. Make these circles the same size as the first. Two new circles, and all the same size.
The new circles will intersect in 2 points. Do you remember the theorem that says this? I hope so, because otherwise you'll have to stop now and prove it. We need those 2 points of intersection,
Notice the nice symmetry we created by using the same size circles! Both the new circles go through point A. (Of course they do. The distance from A to the point J is one radius; the distance back from J to A is also one radius. Same for K. What matters is that we used the same radius for all three circles.)
On the other side of line r, the circles also intersect. And it looks like a mirror image. Is it? Is that the point we want? Yes.
Figure 2.
If you have to prove that these 2 circles will intersect at 2 points, here are some hints for your proof.
Remember that J and K were both on the first circle, so the farthest apart they could be is one diameter (2 radii) – and that would only happen if they were in a line with A. Which they are not, since A is not on r.
Create triangles using the centers
and the intersection points.
You may need to use an indirect proof
(suppose that there were a third
point of intersection …
).
It could be that J and K are closer than 1 radius, but the theorem is still valid.
The point we want is on a line that runs through A and is perpendicular to the line r (by the definition of reflection, remember). Also, the distance from A to the line r must be the same as the distance from the reflected point to r. Let's look at our points as vertices of triangles instead of intersections of circles. (It's just a different way to think about the same points we've been looking at.) Figure 2a shows this.
The line AA′ is perpendicular to r. (We know this from an earlier proof, but if you've forgotten it see the lemma.) So far, so good.
The distances to the line r must also be equal. If you previously proved the theorem that the bisector of the apex angle of an isosceles triangle also bisects the base of the isosceles triangle, that would establish that AP = PA′ Or, we can use our lemma again. It proves the congruence of triangles APJ and A′PJ (and APK and A′PK, too). That in turn implies the congruence of A′P and AP.
Figure 2a.
A Nice, Strong Lemma
Look at the triangles JAK and KA′J. Remember that JA, JA′, KA, and KA′ are all equal because they were created as radii of equal-sized circles. So these are both isosceles triangles. That means the base angles along line r are congruent: ∠KJA ≅ ∠JKA and ∠KJA′ ≅ ∠JKA′.
Besides that,
AA′
≅ A′A
by identity.
So all the corresponding sides
are congruent and the triangles
are congruent to each other [SSS
].
That means ∠JKA ≅ ∠JKA′
and ∠KJA′ ≅ ∠KJA.
Taking both results together, ∠KJA ≅ ∠JKA ≅ ∠KJA′ ≅ ∠JKA′.
Next, look at the other 2 triangles, KAA′ and JAA′. The same logic works for these; they are also isosceles and they are also congruent by SSS. So, ∠JA′A ≅ ∠JAA′ ≅ ∠KAA′ ≅ ∠KA′A.
Finally, look at the smaller triangles.
For triangles AJP and AKP,
JA
≅ KA
because they started
out as radii of equal-sized circles.
We proved that ∠AJP ≅ ∠AKP
and ∠PAK ≅ ∠PAJ.
(OK, we used different names for the angles.
But they are the same angles.)
So AJP ≅ AKP by ASA
.
For A′JP ≅ A′KP the argument is the identical. The argument is almost the same for AJP ≅ A′JP and AKP ≅ A′KP. So, all the little triangles are congruent to each other. This helps us in 2 ways:
- Because corresponding parts are congruent, ∠KPA ≅ ∠JPA ≅ ∠KPA′ ≅ ∠JPA′. If adjacent angles are equal, they are right angles (because they must add to 180°). That establishes that AA′ is perpendicular to line r.
- Because corresponding parts are congruent, AP = PA′. That establishes that A and A′ are the same distance from line r.
All that work, and so far we managed to reflect just the one point A. I can see how this might seem discouraging if it weren't so much fun.
Now that we know how to do it, we can easily reflect each of the vertices around line r. Figures 3 and 4 show how we find the reflection of point D.
First establish 2 points on r; I called them L and M this time. Then draw 2 circles centered at L and at M to find the reflection of point D.
Remember how using the same radius for all 3 circles let us find the line perpendicular to r (the line DD′) and the correct point on that line (D′) in a single step.
Figure 3.
Figure 4.
Repeat the process for points B and C.
Now if you connect the vertices you will have the entire reflected quadrilateral.
Figure 5.
Oh, oh! That last section contains a statement without any support. Did that bother you? Good! That's means you are thinking like a mathematician. You should never let your textbook or your teacher off the hook. Always demand careful reasoning for every claim.
The claim is that the non-vertex points on the sides of the quadrilateral ABCD will lie on the sides of the new quadrilateral A'B'C'D'. It seems obvious that this should be so, but how can we be sure?
Well, maybe you can let your teacher slide a little bit once in a while. But not too often.
