Basic Proportionality Theorem (Thales' Theorem)

The Theorem (BPT / Thales' Theorem)

Here is one of the most useful theorems in geometry, discovered by the Greek mathematician Thales.

Basic Proportionality Theorem (BPT): If a line is drawn parallel to one side of a triangle to intersect the other two sides at distinct points, then it divides those two sides in the same ratio.

In $\triangle ABC$, if a line parallel to $BC$ meets $AB$ at $D$ and $AC$ at $E$, then:

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

Key Point: The line must be parallel to the third side. Then the two sides it cuts are split in equal ratios.

[Board Important] BPT is proved using the areas of triangles on the same base between the same parallels. Learn the statement precisely — it is the foundation of the whole chapter.

Equivalent Ratio Forms

The BPT ratio $\dfrac{AD}{DB} = \dfrac{AE}{EC}$ can be rearranged into other useful forms:

• $\dfrac{AD}{AB} = \dfrac{AE}{AC}$ (each small part to the whole side)
• $\dfrac{DB}{AD} = \dfrac{EC}{AE}$ (taking reciprocals)
• $\dfrac{AB}{DB} = \dfrac{AC}{EC}$

These follow by simple algebra (adding 1, inverting, etc.).

Key Point: Once you have one BPT ratio, you can convert to whichever form the problem needs. The most common in problems is $\dfrac{AD}{DB} = \dfrac{AE}{EC}$.

[Board Important] Be careful which segments are 'parts' and which are 'whole'. $\dfrac{AD}{DB}$ (part : part) is different from $\dfrac{AD}{AB}$ (part : whole).

The Converse of BPT

The converse is equally important and is used to prove lines are parallel.

Converse of BPT: If a line divides two sides of a triangle in the same ratio, then the line is parallel to the third side.

So in $\triangle ABC$ with $D$ on $AB$ and $E$ on $AC$: if $\dfrac{AD}{DB} = \dfrac{AE}{EC}$, then $DE \parallel BC$.

Key Point: Use the BPT to find lengths/ratios when a line is known to be parallel; use the converse to prove parallelism when the ratios are equal.

[Board Important] A typical exam line: 'Show that $DE \parallel BC$.' Compute the two ratios; if they are equal, cite the converse of BPT to conclude parallelism.

Applications — Trapezium and Beyond

BPT appears in trapeziums and in problems with several parallel lines.

Trapezium application

In a trapezium $ABCD$ with $AB \parallel DC$, the diagonals intersect at $O$. Triangles formed there give $\dfrac{AO}{OC} = \dfrac{BO}{OD}$ (the diagonals cut each other in the same ratio).

Key Point: Whenever a line is parallel to a side (or to the parallel sides of a trapezium), set up the equal-ratio relation from BPT.

[Board Important] In a trapezium with $AB \parallel DC$, the diagonals divide each other proportionally — a frequent 3-mark proof or computation.

Solved Examples

Example 1: Find a length using BPT

In $\triangle ABC$, $DE \parallel BC$ with $D$ on $AB$, $E$ on $AC$. If $AD = 2$ cm, $DB = 3$ cm, $AE = 4$ cm, find $EC$.

Solution:

1. By BPT, $\dfrac{AD}{DB} = \dfrac{AE}{EC}$.
2. $\dfrac{2}{3} = \dfrac{4}{EC} \Rightarrow EC = \dfrac{4 \times 3}{2} = 6$ cm.

Final Answer: $EC = 6$ cm.

Takeaway: Substitute into $\dfrac{AD}{DB} = \dfrac{AE}{EC}$ and cross-multiply.

Example 2: Find $x$

In $\triangle ABC$, $DE \parallel BC$. $AD = x$, $DB = x - 2$, $AE = x + 2$, $EC = x - 1$. Find $x$.

Solution:

1. BPT: $\dfrac{x}{x-2} = \dfrac{x+2}{x-1}$.
2. Cross-multiply: $x(x-1) = (x+2)(x-2) \Rightarrow x^2 - x = x^2 - 4$.
3. $-x = -4 \Rightarrow x = 4$.

Final Answer: $x = 4$.

Takeaway: BPT often produces a solvable equation in $x$.

Example 3: Prove $DE \parallel BC$

In $\triangle ABC$, $D$ on $AB$, $E$ on $AC$ with $AD = 1.5$ cm, $DB = 3$ cm, $AE = 1$ cm, $EC = 2$ cm. Is $DE \parallel BC$?

Solution:

1. $\dfrac{AD}{DB} = \dfrac{1.5}{3} = \dfrac{1}{2}$; $\dfrac{AE}{EC} = \dfrac{1}{2}$.
2. The ratios are equal, so by the converse of BPT, $DE \parallel BC$.

Final Answer: Yes, $DE \parallel BC$.

Takeaway: Equal ratios ⇒ parallel (converse of BPT).

Example 4: Part-to-whole form

In $\triangle PQR$, $ST \parallel QR$ with $S$ on $PQ$, $T$ on $PR$. If $PS = 3$, $PQ = 8$, $PT = 4.5$, find $PR$.

Solution:

1. By BPT (part to whole): $\dfrac{PS}{PQ} = \dfrac{PT}{PR}$.
2. $\dfrac{3}{8} = \dfrac{4.5}{PR} \Rightarrow PR = \dfrac{4.5 \times 8}{3} = 12$.

Final Answer: $PR = 12$.

Takeaway: Use the part-to-whole form when whole sides are given.

Example 5: Trapezium diagonals

In trapezium $ABCD$ with $AB \parallel DC$, the diagonals meet at $O$. If $AO = 3$, $OC = 6$, $BO = 4$, find $OD$.

Solution:

1. Diagonals divide each other in the same ratio: $\dfrac{AO}{OC} = \dfrac{BO}{OD}$.
2. $\dfrac{3}{6} = \dfrac{4}{OD} \Rightarrow OD = \dfrac{4 \times 6}{3} = 8$.

Final Answer: $OD = 8$.

Takeaway: Trapezium diagonals split proportionally.

Example 6: Mid-point line

In $\triangle ABC$, $D$ is the midpoint of $AB$ and $DE \parallel BC$ meets $AC$ at $E$. Show $E$ is the midpoint of $AC$.

Solution:

1. $D$ midpoint ⇒ $AD = DB$, so $\dfrac{AD}{DB} = 1$.
2. By BPT, $\dfrac{AE}{EC} = \dfrac{AD}{DB} = 1 \Rightarrow AE = EC$.
3. So $E$ is the midpoint of $AC$.

Final Answer: $E$ is the midpoint of $AC$.

Takeaway: This is the midpoint theorem, a special case of BPT.

Example 7: Find $DB$

In $\triangle ABC$, $DE \parallel BC$. $AD = 4$ cm, $AE = 8$ cm, $EC = 12$ cm. Find $DB$.

Solution:

1. $\dfrac{AD}{DB} = \dfrac{AE}{EC} \Rightarrow \dfrac{4}{DB} = \dfrac{8}{12} = \dfrac{2}{3}$.
2. $DB = \dfrac{4 \times 3}{2} = 6$ cm.

Final Answer: $DB = 6$ cm.

Takeaway: Rearrange the proportion to solve for the unknown segment.

Example 8: Two unknown ratios

In $\triangle ABC$, $DE \parallel BC$. If $\dfrac{AD}{DB} = \dfrac{3}{4}$ and $AC = 14$ cm, find $AE$ and $EC$.

Solution:

1. By BPT, $\dfrac{AE}{EC} = \dfrac{3}{4}$, so $AE : EC = 3 : 4$.
2. $AE + EC = AC = 14$. Parts: $AE = \dfrac{3}{7}(14) = 6$, $EC = \dfrac{4}{7}(14) = 8$.

Final Answer: $AE = 6$ cm, $EC = 8$ cm.

Takeaway: Split the whole side in the BPT ratio.

Example 9: Not parallel

In $\triangle ABC$, $D$ on $AB$, $E$ on $AC$ with $AD = 2$, $DB = 4$, $AE = 3$, $EC = 5$. Is $DE \parallel BC$?

Solution:

1. $\dfrac{AD}{DB} = \dfrac{2}{4} = \dfrac{1}{2}$; $\dfrac{AE}{EC} = \dfrac{3}{5}$.
2. $\dfrac{1}{2} \neq \dfrac{3}{5}$.

Final Answer: No, $DE$ is not parallel to $BC$.

Takeaway: Unequal ratios ⇒ not parallel (converse fails).

Example 10: BPT with algebra

In $\triangle ABC$, $DE \parallel BC$ with $AD = 4x - 3$, $DB = 3x - 1$, $AE = 8x - 7$, $EC = 5x - 3$. Find $x$.

Solution:

1. BPT: $\dfrac{4x-3}{3x-1} = \dfrac{8x-7}{5x-3}$.
2. Cross-multiply: $(4x-3)(5x-3) = (8x-7)(3x-1)$.
3. $20x^2 - 27x + 9 = 24x^2 - 29x + 7 \Rightarrow 4x^2 - 2x - 2 = 0 \Rightarrow 2x^2 - x - 1 = 0$.
4. $(2x + 1)(x - 1) = 0 \Rightarrow x = 1$ (reject $-\tfrac{1}{2}$ as it makes lengths negative).

Final Answer: $x = 1$.

Takeaway: Cross-multiplying BPT can give a quadratic; reject inadmissible roots.