Geometry 2.22 Basic Proportionality Theorem or Thale's Theorem

A straight line drawn parallel to a side of a triangle, divides the other two sides proportionally.
Data: Triangle ABC, D is a point on AB and the line DE is parallel to BC
To prove: AD/BD = BE/CE

Consider triangle ABC and triangle BDE

Step

Statement

Reason

1

ang BAC = ang BDE

Corresponding angles because DE ll AC

2

ang BCA = ang BED

Corresponding angles because DE ll AC

3

ang DBE = ang ABC

Common angle

4

triangle ABC lll triangle BDE

AAA postulate on similarity

5

AB/BD = BC/BE

corresponding sides are proportional

6

(BD+DA)/BD =(BE+EC)/BE

Split AB and AC

7

1+AD/BD = 1+CE/BE

Simplification

8

AD/BD = CE/BE

Subtract 1 from either sides

10

AD/BD= BE/CE

Reciprocating

Converse
We shall try to prove the converse of basic proportionality theorem.
Let us consider triangle ABC shown in the figure below.

Divide the sides of the triangle AB and AC into any number of equal parts.
By measuring the angles B_{1} and B we observe that they are equal.
But B_{1} and B are corresponding angles.
Hence B_{1}C_{1} ll BC.
Similarly
AB_{2}/B_{2} B= AC_{2}/C_{2}C = 2/3 and
B_{2}2C_{2} ll BC
AB_{3}/B_{3}B =AC_{3}/C_{3}C= 3/2 and B_{3}C_{3} ll BC
AB_{4}/B_{4}B = AC_{4}/C_{4}C and B_{4}C_{4} ll BC

Example: In the figure, find the length of PS given that ST ll QR.

Let PS = x cm
As ST ll QR
PS/QS = RT/PT
x/3 = 3/2
x = 9/2 Answer: PS = 4.5 cms