Thales' Theorem
Senior 2 · Mathematics
Thales' Theorem
Lesson 3 of 4
22 min
Thales' idea does not stop at triangles. Parallel lines cut any two crossing lines in proportional pieces: you can see it in the rungs of a ladder, the parallel shelves of a bookcase, and the lines of an exercise book. The converse lets you prove that two lines are parallel just by measuring lengths, which is how builders check their work.
If three or more parallel lines cut two other lines (transversals), the pieces on one transversal are in the same ratio as the corresponding pieces on the other.
Let ABCD be a trapezium with AB ∥ DC. Draw EF parallel to both bases, with E on AD and F on BC. Then
AE/ED = BF/FC.
Why: draw the diagonal AC, crossing EF at G. In triangle ADC, EG ∥ DC, so AE/ED = AG/GC. In triangle CAB, GF ∥ AB, so CG/GA = CF/FB, which means AG/GC = BF/FC. Both ratios equal AG/GC, so AE/ED = BF/FC.
The same diagonal lets you find the length of EF: EF = EG + GF, where each piece comes from one triangle.
Worked example 1
AB = 4 cm, DC = 14 cm, AE = 2 cm and ED = 3 cm, so AD = 5 cm.
Special case: the midline. If E and F are the midpoints of the legs, EF = (AB + DC) ÷ 2. For AB = 6 cm and DC = 10 cm, EF = 8 cm.
In triangle ABC with D on AB and E on AC: if AD/DB = AE/EC, then DE ∥ BC.
Worked example 2
AD = 3, DB = 4.5, AE = 4 and EC = 6. Then AD/DB = 3 ÷ 4.5 = 0.667 and AE/EC = 4 ÷ 6 = 0.667. The ratios are equal, so DE ∥ BC.
Worked example 3
AD = 2, DB = 3, AE = 3 and EC = 4. Now 2 ÷ 3 = 0.667 but 3 ÷ 4 = 0.75. The ratios differ, so DE is not parallel to BC.
Tip
Exam technique. For a trapezium, do not guess that EF is the average of the bases: that is true only for the midline. Draw a diagonal and use Thales twice, once in each triangle. To prove lines parallel with the converse, compute both ratios as fractions in lowest terms (34.5 = 23 and 46 = 23) and state "equal ratios, so DE ∥ BC (converse of Thales' theorem)".
Three parallel lines cut one transversal into pieces of 3 cm and 4.5 cm. On a second transversal the first piece is 4 cm. Find the second piece.
Solution: 34.5 = 4/x, so 3x = 18 and x = 6 cm.
Trapezium ABCD has AB ∥ DC, AB = 6 cm and DC = 10 cm. E and F are the midpoints of AD and BC. Find EF.
Solution: EF = (6 + 10) ÷ 2 = 8 cm.
In trapezium ABCD (AB ∥ DC), EF ∥ AB with AE = 4, ED = 6 and BF = 5. Find FC.
Solution: AE/ED = BF/FC gives 46 = 5/FC, so FC = 7.5.
In triangle PQR, S is on PQ and T on PR with PS = 5, SQ = 10, PT = 4 and TR = 8. Is ST ∥ QR?
Solution: 510 = 12 and 48 = 12. Equal ratios, so ST ∥ QR by the converse.
Remember
Sources and credits
The Rwandan worked examples, the practice items and the exercises were written by Ganzaa.
