Thales' Theorem
Senior 2 · Mathematics
Thales' Theorem
Lesson 2 of 4
24 min
The midpoint theorem is a special case. What if the line parallel to BC cuts the sides not at their midpoints, but one third or two fifths of the way along? Thales' theorem answers this for every position. It is named after Thales of Miletus, a Greek mathematician who, according to tradition, used this idea about 2,600 years ago to measure the height of an Egyptian pyramid from its shadow.
It is the basis of scale drawings, maps, photography and every problem where you find a length you cannot reach from lengths you can.
Thales' theorem (in a triangle). In triangle ABC, let D lie on AB and E on AC. If DE ∥ BC, then
AD/AB = AE/AC = DE/BC and also AD/DB = AE/EC.
In words: a line parallel to one side of a triangle divides the other two sides in the same ratio.
Because DE ∥ BC, angle ADE = angle ABC and angle AED = angle ACB (corresponding angles), and the angle at A is shared. So triangle ADE is an enlargement of triangle ABC with centre A: the two triangles are similar, and all their corresponding sides are in the same ratio. That gives AD/AB = AE/AC = DE/BC. Taking the parts that are left (DB = AB − AD and EC = AC − AE) gives AD/DB = AE/EC.
The midpoint theorem is the case where the ratio is ½.
DE ∥ BC, AD = 4 cm, DB = 6 cm and AE = 5 cm. Find EC.
AD/DB = AE/EC gives 46 = 5/EC. Cross-multiply: 4 × EC = 30, so EC = 7.5 cm.
DE ∥ BC, AD = 3 cm, AB = 9 cm and BC = 12 cm. Find DE.
Here you must use the whole side: DE/BC = AD/AB = 39. So DE = 12 × 39 = 4 cm.
Common error: using AD/DB = 36 gives 6 cm, which is wrong, because DE and BC are sides of the small and the large triangle, so they must be compared with AD and AB, not with DB.
DE ∥ BC, AD = x, DB = 6, AE = 4 and EC = 8. Then x/6 = 48, so x = 3.
Tip
Memory hook: "small over big" or "top over bottom". Either compare the small triangle with the big one (AD/AB = AE/AC = DE/BC), or compare the top part of each side with its bottom part (AD/DB = AE/EC). Never mix the two in one equation, and never put the parallel segment DE into a "top over bottom" ratio.
DE ∥ BC, AD = 6 cm, DB = 4 cm and AE = 9 cm. Find EC.
Solution: 64 = 9/EC, so 6 × EC = 36 and EC = 6 cm.
DE ∥ BC, AB = 10 cm, AD = 4 cm and DE = 6 cm. Find BC.
Solution: DE/BC = AD/AB, so 6/BC = 410 and BC = 15 cm.
DE ∥ BC with AD : AB = 2 : 5 and BC = 15 cm. Find DE.
Solution: DE = 15 × 25 = 6 cm.
DE ∥ BC, AD = x + 1, DB = 4, AE = 6 and EC = 8. Find x.
Solution: (x + 1)/4 = 68, so x + 1 = 3 and x = 2.
Remember
Sources and credits
The Rwandan worked examples, the practice items and the exercises were written by Ganzaa.
