the side-splitter theorem
/ TAY-leez (for Thales) /
Picture a triangle and slice across it with a line drawn parallel to one of its sides. That parallel line cuts the other two sides into pieces — and the side-splitter theorem says it cuts them proportionally. The two upper pieces have the same ratio as the two lower pieces. A line parallel to the base divides the legs in a perfectly balanced way.
In triangle ABC, let a line meet side AB at D and side AC at E, with DE parallel to BC. Then AD/DB = AE/EC. Using the addition property you can also write it as AD/AB = AE/AC = DE/BC, comparing each upper piece (and the cut itself) to the whole. The reason is similar triangles: because DE is parallel to BC, triangle ADE has the same angles as triangle ABC (AA), so corresponding sides are proportional, and the equal ratios fall out. The converse is just as useful: if a line cuts two sides of a triangle in equal ratios, then that line must be parallel to the third side — a clean test for parallelism using only lengths.
This theorem is the bridge between parallels and proportion, and it does real work: it lets you divide a segment into any number of equal parts with straightedge and compass, it underlies the construction of a fourth proportional, and its converse proves lines parallel without measuring a single angle. The name 'Thales' intercept theorem' honours the ancient Greek who, by legend, used exactly this idea to measure the height of the Great Pyramid from its shadow.
In triangle ABC, DE is parallel to BC with AD = 4, DB = 6, and AE = 6. Then AD/DB = AE/EC gives 4/6 = 6/EC, so EC = 9. The parallel cut splits both legs in the ratio 2 : 3.
A line parallel to one side splits the other two sides in equal ratios.
The theorem needs the cutting line to be parallel to the third side; a line that merely crosses both sides at random does not split them proportionally. And note DE/BC equals AD/AB, not AD/DB — the segment-to-base ratio compares the cut to the whole side, not to the leftover piece.