Assignment # 6

Constructing Equal Segments

by

Michael B. Ferra

Problem

Given three points A, B and C. Construct a line intersecting AC in the point X and BC in the point Y such that

AX = XY = YB

Prove your construction is valid by means mathematical proof.

Solution to Problem

STEP 1: Given three points A, B and C. Construct these points to create two line segments which will be defined here as segment AC and BC. These two segments share the point C, and have thus created a resulting angle between the two segments. This angle will be defined as ∠ACB. For the sake of this problem, I am also going to show the construction of a segment AB, which I will show using a dashed segment.

STEP 2: Plot point X' on AC. Construct a circle with center X' (which we will refer to as circle X') and radius AX'.

STEP 3: Now construct a circle with center B (which we will refer to as circle B) having radius AX'. In doing this we have created two congruent circles. The point of intersection of this new circle and segment BC will labeled as Y'. Since Y' is a point on circle B, the segment Y'B can be defined as a radius of this circle. Since we already know circle B has a radius of AX', then the distance from B to any point on this circle will also have this defined radius. We can thus conclude AX' = Y'B.

STEP 4: Construct a parallel line to AB going through the point Y'. This parallel line will intersect circle X' twice but for our focus, we are only concerned with the intersection point that lies within ∠ACB, which will be labeled as Y''. Because Y'' lies on circle X', we can deduce X'Y'' forms its radius. We already know AX' is a radius of circle X', thus AX' = X'Y''.

STEP 5: Construct a parallel line to CB going through Y''. This will intersect AB at a point which will be labeled B'.

*NOTE: Observe in step 5 that a quadrilateral can be created by connecting points B, Y', Y'', and B'. Now let's restate what we've done in steps 4 and 5. We first constructed a parallel line to AB through Y' creating point Y''. We then followed by constructing a parallel line to CB through Y'' creating point B'. Notice from the constructions in previous steps that quadrilateral BY'Y''B' is a parallelogram. Thus, we can deduce Y''B' = Y'B. If we combine this with our previous results then we can now state AX' = X'Y'' = Y''B'. We have now created the result of the segments we were looking for, yet on a similar, smaller scale. The only thing that remains is projecting these points X' and Y'' onto our segments AC and CB such that we maintain this similarity. Before continuing to step 6, let's see a graph that's a little cleaner and finish from there.

STEP 6: Construct a line through A and Y'' that intersects segment BC. By doing this, we are projecting Y'' onto segment CB, giving us the point Y we are solving for, thus this point of intersection will be labeled Y.

STEP 7: In order to maintain similarity in finding our point X, we must create a line parallel to X'Y'' through our new point Y. This parallel line will intersect line AC at a point which will define our point X, thus it is labeled as such.

Once again, let's clean this up, showing only points A, B, C, X, and Y. Doing this will provide us with our final result such that AX = XY = YB.

I have used Geometer's Sketchpad (GSP) to create these illustrations but by no means do you need this to solve this problem. Given the instructions I've laid out, paper and pencil will substitute just fine. Since I do have GSP at my disposal, I want to show one final graph to show that my segments have come out equal. By using the measure tab, I can click two points and find the distance between them. As you can see for this example I've created, AX = XY = YB = 5.50 cm, thus the steps I have taken to get to my final solution are verified. This is showing a specific example so if you'd like to explore more possibilities, you can click the link below the diagram.