9 min readPathway sheetOften a 3-mark or 5-mark proof; the result is used in 2-mark problems

Basic Proportionality Theorem (with proof)

If a line drawn parallel to one side of a triangle cuts the other two sides at distinct points, it divides those sides in the same ratio.

Statement in symbolsIn ΔABC, if DE ∥ BC with D on AB and E on AC, then AD/DB = AE/EC

Also calledThales theorem

The proof usesArea of a triangle = ½ × base × height; triangles on the same base and between the same parallels have equal areas

Other useful formsAD/AB = AE/AC and DB/AB = EC/AC

How to draw the figure for the proof

Draw a triangle ABC with the base BC at the bottom.
Mark D on AB and E on AC, and draw DE parallel to BC. Write DE ∥ BC.
Join BE and CD (dotted lines are fine).
From E, draw EN ⊥ AB with N on AB. From D, draw DM ⊥ AC with M on AC.
Label every point: A, B, C, D, E, M and N.

The proof, step by step

Given: in ΔABC, DE ∥ BC, with D on AB and E on AC. To prove: AD/DB = AE/EC. Construction: join BE and CD; draw DM ⊥ AC and EN ⊥ AB.

1
ar(ADE) = ½ × AD × EN and ar(BDE) = ½ × DB × EN
EN is the height of both triangles
2
So ar(ADE)/ar(BDE) = AD/DB ... (1)
3
ar(ADE) = ½ × AE × DM and ar(DEC) = ½ × EC × DM
DM is the height of both triangles
4
So ar(ADE)/ar(DEC) = AE/EC ... (2)
5★
ΔBDE and ΔDEC are on the same base DE and between the same parallels DE and BC. So ar(BDE) = ar(DEC) ... (3)
your textbook uses this area fact without proving it, so just quote it as the reason
6
From (1), (2) and (3), AD/DB = AE/EC. Hence proved.

★ key step

Why the heights are the same

Triangles ADE and BDE have their bases AD and DB on the same line AB. Their third corner is E in both. So the perpendicular EN from E to AB is the height of both. In the same way, DM is the height of both ADE and DEC. When you divide the two areas, ½ and the height cancel. Only the ratio of the bases is left.

Example 1: using BPT

Problem

In ΔABC, DE ∥ BC with D on AB and E on AC. AD = 3 cm, DB = 4.5 cm and AE = 4 cm. Find EC.

Working

DE ∥ BC, so by BPT, AD/DB = AE/EC
3/4.5 = 4/EC
EC = (4 × 4.5)/3 = 18/3 = 6 cm

Answer

EC = 6 cm

Example 2: BPT with algebra

Problem

In ΔPQR, ST ∥ QR with S on PQ and T on PR. PS = x + 1, SQ = x − 1, PT = x + 5 and TR = x + 1 (in cm). Find x.

Working

ST ∥ QR, so by BPT, PS/SQ = PT/TR
(x + 1)/(x − 1) = (x + 5)/(x + 1)
Cross-multiply: (x + 1)² = (x + 5)(x − 1)
x² + 2x + 1 = x² + 4x − 5
6 = 2x, so x = 3
Check: PS = 4, SQ = 2, PT = 8, TR = 4, and 4/2 = 8/4 = 2.

Answer

x = 3

In real life

The crossbar of an A-frame step ladder is parallel to the floor. So it cuts both legs in the same ratio, which is BPT at work.

Exam angle

5 marks: “State and prove the Basic Proportionality Theorem.” State BPT. Draw ΔABC with DE ∥ BC, join BE and CD, draw EN ⊥ AB and DM ⊥ AC. Write ratios of areas (1) and (2), equal areas ar(BDE) = ar(DEC) (3), and conclude AD/DB = AE/EC.

2 marks: “In ΔABC, DE ∥ BC with D on AB and E on AC. If AD/DB = 2/5 and AC = 4.9 cm, find AE.” By BPT, AE/EC = 2/5, so AE is 2 parts out of 7. AE = 2/7 × 4.9 = 1.4 cm.

3 marks: “In ΔABC, D, E and F lie on AB, AC and BC, with DE ∥ BC and EF ∥ AB. Prove that AD/DB = BF/FC.” DE ∥ BC gives AD/DB = AE/EC (BPT in ΔABC). EF ∥ AB gives CE/EA = CF/FB, so AE/EC = BF/FC. Hence AD/DB = BF/FC.

Remember it

Parallel line, proportional sides. For the proof: two heights, one same base, equal areas.