OddClasses — Odd ones figure it out.

The Proportion Theorem (Basic Proportionality)

Theorem 1📌 Learn for exam

Proportion Theorem

If a line is drawn parallel to one side of a triangle, it divides the other two sides proportionally.

If $DE \parallel BC$ with $D$ on $AB$ and $E$ on $AC$ , then:

\frac{AD}{DB} = \frac{AE}{EC}

Also written: $\dfrac{AD}{AB} = \dfrac{AE}{AC} = \dfrac{DE}{BC}$

▸ Proportion Theorem — drag D along ABwatch AD/DB = AE/EC stay equal as you drag

Loading diagram…

Converse of Proportion Theorem

\dfrac{AD}{DB} = \dfrac{AE}{EC}

, then

DE \parallel BC

Theorem 2📌 Learn for exam

Proportion Theorem in Practice

If $DE \parallel BC$ , then:

\frac{DE}{BC} = \frac{AD}{AB} = \frac{AE}{AC}

Consequence: $\triangle ADE \;|||\; \triangle ABC$ (AA similarity — all corresponding sides in proportion).

Show proof ▾

Given

DE \parallel BC

;

D

AB

E

AC

Draw heights

Draw

h_1

from

D \perp AE

and

h_2

from

E \perp AD

. Join

DC

and

BE

Area ratios

\dfrac{\text{Area}(\triangle ADE)}{\text{Area}(\triangle BDE)} = \dfrac{AD}{DB}

— triangles with same base

DE

, heights from

A

and

B

Similarly

\dfrac{\text{Area}(\triangle ADE)}{\text{Area}(\triangle CED)} = \dfrac{AE}{EC}

Key fact

\text{Area}(\triangle BDE) = \text{Area}(\triangle CED)

— same base

DE

, same height between parallel lines

DE

and

BC

Therefore

$\dfrac{AD}{DB} = \dfrac{AE}{EC} \checkmark$

Tip

When you see DE ∥ BC in a question, immediately write

\frac{AD}{DB} = \frac{AE}{EC}

. This is usually the first step in the proof.

Worked Example

In △ABC, DE ∥ BC with D on AB and E on AC. AD = 4, DB = 6, AE = 5. Find EC.

Worked Example

In △PQR, ST ∥ QR with S on PQ and T on PR. PS = 3x, SQ = 12, PT = x + 2, TR = 8. Find x.