Basic Proportionality Theorem (Thales)

The single most useful proportionality result in this chapter is the Basic Proportionality Theorem, often called Thales' Theorem in Indian textbooks.

Statement and proof

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

Setup. In $\triangle ABC$ , let $DE \parallel BC$ with $D$ on $AB$ and $E$ on $AC$ . Then $\frac{AD}{DB} = \frac{AE}{EC}.$

Proof (using areas). Join $BE$ and $CD$ .

Triangles $\triangle ADE$ and $\triangle BDE$ share the base $DE$ and have heights from $A$ and $B$ to $DE$ . So $\frac{[\triangle ADE]}{[\triangle BDE]} = \frac{AD}{DB} \quad (\text{areas in ratio of bases on the same line}).$ Wait , both triangles have base $DE$ but different heights. Actually a cleaner version: $\triangle ADE$ and $\triangle BDE$ share vertex $E$ and have bases $AD, DB$ on the same line $AB$ . Heights from $E$ to $AB$ are equal. So $\frac{[\triangle ADE]}{[\triangle BDE]} = \frac{AD}{DB}.$ Similarly, $\triangle ADE$ and $\triangle CED$ share vertex $D$ and have bases $AE, EC$ on $AC$ . Heights from $D$ to $AC$ are equal: $\frac{[\triangle ADE]}{[\triangle CED]} = \frac{AE}{EC}.$ Now $\triangle BDE$ and $\triangle CED$ share base $DE$ . Since $DE \parallel BC$ , both $B$ and $C$ are at equal perpendicular distance from $DE$ . So $[\triangle BDE] = [\triangle CED]$ .

Therefore $[\triangle ADE]/[\triangle BDE] = [\triangle ADE]/[\triangle CED]$ , giving $\frac{AD}{DB} = \frac{AE}{EC}. \qquad \blacksquare$

Converse of BPT

Theorem (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.

Setup. In $\triangle ABC$ , $D$ on $AB$ and $E$ on $AC$ such that $\dfrac{AD}{DB} = \dfrac{AE}{EC}$ . Then $DE \parallel BC$ .

Sketch of proof. Through $D$ draw a line parallel to $BC$ ; by BPT it cuts $AC$ at some point $E'$ with $AE'/E'C = AD/DB = AE/EC$ . So $E' = E$ . The drawn parallel coincides with $DE$ , hence $DE \parallel BC$ . $\blacksquare$

Using BPT

The theorem and converse together let us:

Find unknown lengths in triangles cut by a line parallel to one side.
Prove two lines are parallel by checking ratios.
Establish many ancillary results (the angle bisector theorem, mid-segment theorem, and the similarity criteria coming up).

In particular, the mid-segment theorem ("the line joining the midpoints of two sides is parallel to the third side and half its length") is an immediate consequence: midpoints give the ratio $1 : 1$ on both sides, so by BPT the segment is parallel.

Worked examples

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

By BPT: $AD/DB = AE/EC \Rightarrow 1.5/3 = 1/EC \Rightarrow EC = 2$ cm.

Example 2. In $\triangle ABC$ , $D$ and $E$ are points on $AB$ and $AC$ such that $AD/DB = AE/EC = 1/3$ . Is $DE \parallel BC$ ?

Yes , by the converse of BPT.

Example 3. In $\triangle ABC$ , $D$ on $AB$ , $E$ on $AC$ , with $AD = 2, DB = 3, AE = 4, EC = 6$ . Is $DE \parallel BC$ ?

$AD/DB = 2/3, AE/EC = 4/6 = 2/3$ . Equal ratios $\Rightarrow$ $DE \parallel BC$ (by converse of BPT).

Example 4. In $\triangle ABC$ , $DE \parallel BC$ . If $AD = x, DB = x - 2, AE = x + 2, EC = x - 1$ , find $x$ .

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

Example 5. Through point $P$ on side $AB$ of $\triangle ABC$ , a line parallel to $BC$ meets $AC$ at $Q$ . Show that the area of $\triangle APQ :$ area of trapezium $BPQC = 1 : 8$ if $AP : PB = 1 : 2$ .

Since $PQ \parallel BC$ and $AP/PB = 1/2$ , we get $AP/AB = 1/3$ . As we'll prove in topic 4, $\triangle APQ \sim \triangle ABC$ with ratio $1/3$ , so areas are in ratio $1/9$ . Hence $[\triangle APQ] : [\triangle ABC] = 1 : 9$ , and $[\triangle APQ] : [BPQC] = 1 : (9 - 1) = 1 : 8$ . ✓

Try it yourself

In $\triangle ABC$ , $DE \parallel BC$ , $AD = 4, DB = 6, AE = 3$ . Find $EC$ .
State the converse of BPT.
In $\triangle ABC$ , $D$ on $AB$ , $E$ on $AC$ , $AD/AB = 1/4$ , $AE/AC = 1/4$ . Is $DE \parallel BC$ ?
In $\triangle ABC$ , $DE \parallel BC$ , $AD = 5.7, DB = 9.5, AE = 4.8$ . Find $EC$ .
In $\triangle ABC$ , $D$ on $AB$ and $E$ on $AC$ such that $DE \parallel BC$ . If $AD = 6, DB = 9$ , and $AE + EC = 25$ , find $AE$ .
State the mid-segment theorem and derive it from BPT.
If a line divides the sides $AB$ and $AC$ of $\triangle ABC$ in the same ratio, prove that the line is parallel to $BC$ .
In $\triangle PQR$ , $S$ on $PQ$ , $T$ on $PR$ , $PS = 4, SQ = 6, PT = 8, TR = 12$ . Is $ST \parallel QR$ ?
The line through the midpoints of two sides of a triangle is parallel to the third , prove via BPT.
In $\triangle ABC$ , $AD$ is the median from $A$ . $E$ is the midpoint of $AD$ . $BE$ meets $AC$ at $F$ . Show that $AF = (1/3) AC$ .

Pitfalls / Insight

BPT is about ratios on the two sides cut, not on the parallel segment itself.
Order matters. $AD/DB$ is not the same as $DB/AD$ , always set up the proportion carefully.
Converse requires the same ratio on the two sides , check before declaring parallel.

Insight. BPT is the bridge between parallel lines and proportionality. Once you see parallel sides in a triangle, jump straight to a ratio equation.