## Basic Proportionality Theorem

**Theorem : **

**If a line parallel to a side of a triangle intersects the remaining sides in two distinct points, then the line divides the sides in the same proportion.**

**Given :**

In ∆ABC,

line l || line BC and line l intersects AB and AC in points P and Q respectively.

**To prove :**

\(\large \frac {AP}{PB}\) = \(\large \frac {AQ}{QC}\)

**Construction: **

Draw seg PC and seg BQ

**Proof :**

∆APQ and ∆PQB have equal heights

∴ \(\large \frac {A (∆APQ)}{A (∆PQB)}\) = \(\large \frac {AP}{PB}\) … (I) [*The ratio of areas of two triangles having equal heights, is equal to the ratio of their corresponding bases*]

and \(\large \frac {A (∆APQ)}{A (∆PQC)}\) = \(\large \frac {AQ}{QB}\) … (II) [*The ratio of areas of two triangles having equal heights, is equal to the ratio of their corresponding bases*]

Now,

seg PQ is the common base of ∆PQB and ∆PQC

and seg PQ || seg BC

Hence ∆PQB and ∆PQC have equal heights

A(∆PQB) = A(∆PQC) ………. (III)

∴ \(\large \frac {A (∆APQ)}{A (∆PQB)}\) = \(\large \frac {A (∆APQ)}{A (∆PQC)}\) … [*from (I), (II) and (III)*]

∴ \(\large \frac {AP}{PB}\) = \(\large \frac {AQ}{QB}\) … [*from (I) and (II)*]

**Hence proved**