Assignment Six: Problem 3


Rekha Payor

Given line segments j, k, m. If these are medians of a triangle, construct the triangle.

Using segments j, k, and m, construct a triangle with these lengths. This can be done using Euclid’s Proposition I. 22. This Proposition is as follows:

First extend segment m into a line (alternatively, you can draw a line and then construct a circle with its center on the line and radius m). Draw a circle with center on the right side of segment m with radius m. On the right side of m also draw a circle with center at the end of m and a radius of k. Where this circle intersects the line, is a segment of length k. At the right end of segment k, draw a circle centered here with radius j. Where this circle intersects the line you have a segment of length j.



Where the circle of radius m and the circle of radius j meet, construct this point. Then connect this point to the right side of segment m and the right side of segment k. This triangle thus has sides of length j, k, and m. Since k is the base, this is clear that one side is equal to k. Since the third point that is in the intersection of circles with radius m and j, we know m and j each are a side of the triangle too.



Now we will rotate our constructed triangle and label the vertices to make the proof easier to follow. Hence we will use triangle ABC to represent the triangle constructed by the given medians:


To construct the triangle with the given medians, we want to create the double of triangle ABC we just created.

First extend line AB. Construct a circle at point A with radius j. Where this circle intersects the extension of line AB call this point D. Draw a line through point D that is parallel to AB. Extend line BC and where this parallel line intersects it call this point E. Draw a line through point A that is parallel to BC. Where this line intersects line DE, call this point F. I claim triangle BDE is the double of triangle BAC.


Notice that since AC is parallel to DE with straight line DB falling on them so angle BAC is equal to angle ADF. This is true since “a straight line falling on parallel straight lines makes the alternative angles equal to one another” (proposition I.28). Similarly since BE is parallel to AF with straight line DB falling on them, angle ABC is equal to angle DAF (proposition I.28). Finally, by construction AB is equal to AD. Hence by angle side angle we have that triangle ABC is congruent to triangle DAF. Thus all remaining corresponding angles and sides are equal. Namely, AC is equal to DF and BC is equal to AF.

Notice that there is a parallelogram ACFE. In parallelogrammic areas, opposite sides are equal so AC is equal to FE and AF is equal to CE (proposition I.34). But we already know BC is equal to AF, which is now equal to CE, so BC is equal to CE. Hence we have bisected BE at point C. Similarly, AC is equal to DF which is now equal to FE, so DF is equal to FE. Thus, DE has been bisected at point F.

Hence we have created triangle BDE from the given medians, since DB is 2j, BE is 2k, and DE is 2m.