(Note: If you want to recreate these graphs as you make your way through the construction, check out this graph on Desmos.com. Click the folder icons for Steps 1 through 4 to make the graphs appear.)
Proposition 2
To place at a given point a straight line equal to a given straight line.
Here is a picture of a point A and a line segment BC.
We are being asked to draw a line segment starting at point A that has the same length as line segment BC. I drew line segment BC to be vertical, but that choice is arbitrary and will not affect the proof. You may be thinking “Why not just measure BC and then draw a line segment starting at A of the same length?” The concept of this construction is that it must be done only using the previous definitions, postulates, common notions, and propositions – without actually measuring.
According to Proposition 1, we know that we can construct an equilateral triangle (ABD) on the line segment AB.
At first glance that may not appear to help, but this is where you must really appreciate the abilities of these ancient mathematicians. Constructing the equilateral triangle will provide us with two points (B and D) that can be the centers of two circles that will allow us to relate the length BC to another line segment starting at A.
According to Postulate 2 we can extend the line segment DA beyond A to another point E and we can extend the line segment DB beyond B to another endpoint F, as shown.
We know that the length BC has to come into play at some point. We can draw a circle, centered at B, that has a radius (or, as Euclid would call it, a distance) equal to BC.
We label the point where the circle intersects line segment DF as G.
The segment DG can be thought of as the sum of DB and BG.
We know that the segment DA is the same as DB (they are two sides in an equilateral triangle), and if we can find a point L on the line segment DE such that the length DL is the same as the length DG, then will show that AL is the same as BG, which we have already shown as being the same as BC, and we will have the line segment we were looking for.
To make this work we will need a circle centered at D, and we will make the radius equal to DG.
The point where this circle intersects the line segment DE is labeled as L.
We know that DL is equal to DG, because L and G are both on the same circle that is centered as D. We also know that DA is the same as DB because those are two sides of an equilateral triangle.
Now we use Common Notion 3. If equals (DA and DB) be subtracted from equals (DL and DG respectively), the remainders (AL and BG) are equal.
Put another way, we had two line segments of equal length (DL and DG) and we cut equal pieces (DA and DB) off of those segments. The remaining segments (AL and BG) must also be equal to each other.
So, AL is equal to BG. But we also know that BG is equal to BC because they were on the same circle centered at B. By Common Notion 1 (Things which are equal to the same thing are also equal to each other), we therefore know that AL is equal to BC.
We have found a line segment beginning at point A that is equal to the given line segment BC.
Being what it was required to do.
As you make your way towards being a mathematician you begin with the final picture and see if you can prove that AL is equal to BC. Once you are able to do that, the next step for you is to see if you can determine the next step if the circle centered at D was not provided – in other words, can you reverse engineer this process. This will make you aware why the circle centered at D was important. You can then restart without either circle being provided, and see if you can justify those two circles, and so on.
This construction is nowhere near obvious, but understanding it will be useful in later proofs as well as in later courses.
Proposition 3
This is another construction.
Given two unequal straight lines, to cut off from the greater a straight line equal to the less.
If you are given two line segments of different lengths, can you cut the segment with greater length so that it is equal to the segment of shorter length? This could be within your reach.
Hint 1: Use Postulate 2 to draw a segment equal to the shorter segment that shares an endpoint with the longer segment.
Hint 2: Draw a circle.
Any questions or comments? Please leave feedback below. – George
Appreciating this project George. I’ve always been impressed what the ‘ancient’ mathematicians did through construction. I hope to continue following along.
Thanks Blair. I’m always trying to show my students how great these mathematicians were, considering the tools they had to work with. Hopefully my students find it motivating. Thanks for reading!