Similarity (With Applications to Maps and Models)

318. The Parallel Divider · BPT: Parallel lines cut sides proportionally

The ratios AD/DB and AE/EC remain equal as D moves along AB.

Basic Proportionality Theorem (Thales): a line drawn parallel to one side of a triangle cuts the other two sides in the same ratio: AD/DB = AE/EC. The converse holds too — equal ratios force the line to be parallel.

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Selina ICSE: Similarity (With Applications to Maps and Models)

What this lesson covers

Try to break it

Drag D along AB. E slides along AC to keep DE parallel to BC, and the ratios AD/DB and AE/EC always come out equal — the Basic Proportionality Theorem (Thales'). Try to drag D so DE tilts away from being parallel to BC; impossible — E tracks D to preserve the parallel.

How you build it

Draw a line parallel to BC and check that AD/DB = AE/EC.

Mark D at (-2, 4) on side AB. D sits at height y = 4, between A (y = 6) and B (y = 0), so AD : DB = 2 : 4 = 1/2.
Mark E at (2, 4) on side AC, at the SAME height (y = 4) as D. E splits AC as AE : EC = 2 : 4 = 1/2 — the same ratio as D.
Join D to E. Because D and E are at the same height, DE is parallel to BC — and the parallel line has cut the two sides in equal ratios: AD/DB = AE/EC = 1/2 (BPT).

The proof, step by step

Prove that a line parallel to one side of a triangle divides the other two sides in the same ratio.

∠ABC = ∠ADE and ∠ACB = ∠AED [Corresponding angles, since DE || BC]
∠BAC = ∠DAE [Common angle to both triangles]
∴ ΔABC ~ ΔADE [AAA Similarity Postulate]
∴ AB/AD = AC/AE ⇒ (AD+DB)/AD = (AE+EC)/AE ⇒ 1 + DB/AD = 1 + EC/AE ⇒ AD/DB = AE/EC

Worked example

In ΔPQR, line ST is drawn parallel to side QR, meeting PQ at S and PR at T. If PS = 4 cm, SQ = 6 cm, and PR = 15 cm, find the length of PT.

By BPT, PS/SQ = PT/TR. Let PT = x, then TR = 15-x. 4/6 = x/(15-x) ⇒ 2/3 = x/(15-x) ⇒ 30-2x=3x ⇒ 5x=30 ⇒ x=6. So PT = 6 cm.

5 cm
6 cm — correct
7 cm
8 cm