8 min readConcept sheetPart III proof (5 marks); Part I and Part II ratio problems

Thales theorem (basic proportionality) and its converse

A line parallel to one side of a triangle cuts the other two sides in the same ratio. The converse proves lines parallel.

Thales theorem (BPT)In ΔABC, if DE ∥ BC, then AD/DB = AE/EC

Also trueAD/AB = AE/AC = DE/BC

ConverseIf AD/DB = AE/EC, then DE ∥ BC

Proof ideaΔADE ~ ΔABC by AA (corresponding angles)

Theorem vs converse

Thales theoremIf a line parallel to one side of a triangle cuts the other two sides, it divides them in the same ratio. So AD/DB = AE/EC.

ConverseIf a line divides two sides of a triangle in the same ratio, then it is parallel to the third side.

Use the theorem to find a length. Use the converse to prove that two lines are parallel. Thales theorem is also called the basic proportionality theorem (BPT).

How to draw the figure and prove the theorem

Figure: draw ΔABC with base BC. Mark D on AB and E on AC so that DE is parallel to BC. Label A, B, C, D and E.

1
Given: in ΔABC, D is on AB and E is on AC, with DE ∥ BC. To prove: AD/DB = AE/EC.
2
In ΔADE and ΔABC, ∠DAE = ∠BAC.
common angle
3
∠ADE = ∠ABC.
corresponding angles, since DE ∥ BC
4
So ΔADE ~ ΔABC. Hence AB/AD = AC/AE.
AA criterion
5
Write AB = AD + DB and AC = AE + EC: (AD + DB)/AD = (AE + EC)/AE.
6
1 + DB/AD = 1 + EC/AE, so DB/AD = EC/AE.
7
Take reciprocals: AD/DB = AE/EC. Hence proved.

Example 1: find x

Problem

In ΔABC, DE ∥ BC with D on AB and E on AC. AD = x + 1, DB = 3x − 1, AE = x + 3 and EC = 3x + 4. Find x.

Working

By Thales theorem, AD/DB = AE/EC.
(x + 1)/(3x − 1) = (x + 3)/(3x + 4)
Cross-multiply: (x + 1)(3x + 4) = (x + 3)(3x − 1)
3x² + 7x + 4 = 3x² + 8x − 3
7x + 4 = 8x − 3, so x = 7
Check: AD = 8, DB = 20, AE = 10, EC = 25. Both 8/20 and 10/25 equal 2/5.

Answer

x = 7

Example 2: a roof frame (word problem)

Problem

A triangular roof frame ABC has a bar DE parallel to the base BC, with D on AB and E on AC. AD = 1.2 m, DB = 1.8 m and AE = 1.4 m. Find EC.

Working

DE ∥ BC, so by Thales theorem, AD/DB = AE/EC.
1.2/1.8 = 1.4/EC
EC = 1.4 × 1.8/1.2
EC = 2.52/1.2 = 2.1

Answer

EC = 2.1 m

Example 3: use the converse

Problem

In ΔPQR, S is on PQ and T is on PR. PS = 3 cm, SQ = 6 cm, PT = 4 cm and TR = 8 cm. Is ST parallel to QR?

Working

PS/SQ = 3/6 = 1/2
PT/TR = 4/8 = 1/2
PS/SQ = PT/TR, so ST divides PQ and PR in the same ratio.
By the converse of Thales theorem, ST ∥ QR.

Answer

Yes, ST ∥ QR.

In real life

The cross-bar of an A-shaped ladder is parallel to the floor, so it cuts both legs in the same ratio. Carpenters use the same idea in a roof truss.

Exam angle

2 marks: “In ΔABC, DE ∥ BC, AD = 3 cm, DB = 4 cm and AE = 4.5 cm. Find EC.” AD/DB = AE/EC gives 3/4 = 4.5/EC. So EC = 4.5 × 4/3 = 6 cm.

5 marks: “State and prove Thales theorem (basic proportionality theorem).” Statement and figure with DE ∥ BC. ΔADE ~ ΔABC by AA, so AB/AD = AC/AE. Split AB and AC, subtract 1, then take reciprocals: AD/DB = AE/EC.

2 marks: “D and E are on AB and AC of ΔABC. AD = 2 cm, DB = 3 cm, AE = 3 cm and EC = 4.5 cm. Is DE ∥ BC?” AD/DB = 2/3 and AE/EC = 3/4.5 = 2/3. The ratios are equal, so DE ∥ BC by the converse of Thales theorem.

Remember it

Parallel gives same ratio; same ratio gives parallel. The first is Thales theorem, the second is its converse.