Let’s begin with three points A, B, C.
Next, we will construct a smaller version of the congruent segments. To begin we will construct a point D on the segment AC. Then construct a line parallel to segment CB through the point D. Now, how can we create a segment congruent to segment AD through the point D on the parallel line?
Now, we would like to construct a segment from point E to segment AB and congruent to segments AD and DE. So let’s construct a circle with center E and radius length DE.
Next, we will construct a rhombus with sides DE and EF.
The segments AD, DG, and GF are all congruent. Since our figure is getting complicated we will hide objects that we don’t need anymore, such as the circles and the rhombus. Then we will construct a line through the point A and G. Label the point of intersection of the new line and segment CB as Y. Now, construct a line parallel to segment DG through the point Y. What triangles are similar?
Since parallel lines we know corresponding angles are congruent. So <ADG@<AXY, <AGD@<AYX, <AGF@<AYB, and <AFG@<ABY. Therefore, DADG~DAXY and DAGF~DAYB because AA~.
Since there are similar triangles we know corresponding sides are proportional, so
By noticing both triangles share sides AG and AY we know
Since AD = DG = GF, we can conclude AX = AY = YB.