Question

Is the following construction a valid way to drop a perpendicular from a point above a line segment?

Let A and B be the endpoints of the line segment. Let C be a point somewhere above the line segment.
1.) Set compass at point A and set it to length AC. Draw the circle.
2.) Set compass at point B and set it to length BC. Draw the circle.
3.) Locate the 2 points of intersection of the 2 circles created in steps 1 & 2. Label these D and E.
4.) DE is perpendicular to AB.

I can provide an image if these steps aren't clear. Thanks.

Answer 1

I realize this construction is normally done with 3 circles, and I am familiar with the 3 circle method. I'm trying to prove to myself that this 2 circle method always works and is valid.

Answer 2

Really interesting proof, working on it.

Answer 3

All right, I had to make sure it was right (plus I started proving it some oddly complicated way, for some stupid reason), so it took some time, but, here:

So, from a line segment \(AB\), and a third point \(P\), we construct two circles with radii \(AP\) and \(BP\), respectively. Each circle must intersect at two points, one defined, the other we denote \(P'\). The point at the intersection of the line-segment \(PP'\) and line-segment \(AB\), we denote as \(C\).

We know that \(AP\) is a radius of circle \(A\), as is \(AP'\), thus \(\angle APP'=\angle AP'P\), and we have now shown that the triangle \(APP'\) is an isosceles triangle. By this, we know that \(AC\) is a perpendicular bisector of \(PP'\) since \(AC\) is the height of such an isosceles triangle from base \(PP'\). By this, since \(AB\) is collinear to \(AC\), we then have that \(PP' \perp AC\implies PP' \perp AB\), which is what we wished to prove.

Answer 4

Thanks for this. I like this approach much better than what I was trying to do, which was unnecessarily complicated as well. The only thing I need some extra convincing on is "we know that AC is a perpendicular bisector of PP′ since AC is the height of such an isosceles triangle from base PP′." I'm not convinced that AC is the height of the isosceles triangle (though it sure looks like it is!). I'm working on proving now that PC = CP'. This would convince me that AC is the height of the triangle.

(Feel free to let me know if I'm going the wrong way with this. I TA for this class, and it's introductory level college geometry, so the students don't need to do rigorous proofs, but they need to understand why constructions work, which means I need to understand all of the details so I can explain to them in a way they understand.)

Answer 5

All right, haha, like a good mathematician, always question 'hand-waving'. But, yes, it is interesting to note that showing this actually requires both triangles, and some additional work.

As a side-note, for a nice way, I think it may be easier to directly show that \(\angle ACP'=\angle ACP\). I'm trying to finish up a proof of this, but I feel you may finish it before I do.

Answer 6

Triangle APB is congruent to triangle AP'B, using Euclid's Side-Side-Side congruence theorem. (AP = AP' by construction (equal radii), BP = BP' by construction (equal radii), and AB = AB (reflexive).

Therefore ∠PAC = ∠P'AC by corresponding parts of congruent triangles. We also have AP = AP' by construction (equal radii) and AC = AC (reflexive). By Euclid's Side-Angle-Side, triangle PAC is congruent to triangle P'AC.

By corresponding parts of congruent triangles, ∠PCA = ∠P'CA.

Since ∠PCP' is a straight line, ∠PCP' = 180 degrees. By inspection, ∠PCA + ∠P'CA = ∠PCP'. Therefore 2∠PCA = 180, by substitution, and ∠PCA = 90 degrees = ∠P'CA.

Therefore ∠PCA is right (as is ∠P'CA) , which implies AC is perpendicular to PP'. Since AC is colinear with AB, PP' is perpendicular to AB. QED.

Answer 7

This proof seems a little complicated. I think I understand now why I teach the 3 circle method of construction instead of this 2 circle method. Let me know if there are logical gaps or perhaps it could be done more simply.

Answer 8

Nice. I was writing it out as such:

We know that triangles \(APC\cong AP'C\) since \(AB=AB\), \(AP'=AP\), and \(BP'=BP\). By this, we know that \(\angle PAC=\angle P'AC\), which implies that triangles \(PAC\) and \(P'CA\) are also congruent, since \(AC=AC\) and \(AP'=AP\), thus, by equal construction, \(\angle P'CA=\angle PCA\). By this, due to vertical angles, all of \(\angle P'CB=\angle PCA=\angle P'CA=\angle PCB\). Thus, all angles are 90 degrees.
QED.

Definitely not as much rigor as your proof, haha, but it's ideally the same thing.

Answer 9

As for the height, which, for some reason, I had forgotten to mention, you could prove it by forcing a single case on to SSA for any pair of adjacent, congruent triangles. But, noticing the larger triangles was much nicer, personally.

Answer 10

Thanks for your help! I see what you're saying about the isosceles triangle height, but I agree with you: I like this final method we've arrived at. Thanks!