Basic Proportionality Theorem

The Basic Proportionality Theorem (BPT), also known as Thales' Theorem, is a fundamental result in geometry that plays a vital role in understanding the properties of triangles. The theorem forms the foundation of many advanced geometric principles and is frequently used in problems related to triangles and proportionality.

1.0What is the Basic Proportionality Theorem?

The Basic Proportionality Theorem states:

"If a line is drawn parallel to one side of a triangle and intersects the other two sides, then it divides those two sides in the same ratio."

In simple terms, if you draw a line inside a triangle that runs parallel to one side of the triangle, this line will divide the other two sides of the triangle in such a way that the corresponding segments are proportional.

2.0Statement of the Basic Proportionality Theorem

Given a triangle ABC, if a line DE is drawn parallel to side BC and intersects sides AB and AC at points D and E, respectively, then:

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

This means that the ratio of the divided segments on side AB is equal to the ratio of the divided segments on side AC.

3.0Proof of the Basic Proportionality Theorem

Given:

In triangle ABC, DE is a line parallel to side BC, and it intersects sides AB and AC at points D and E, respectively.

To Prove:

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

Construction:

Join points B and E, and C and D to form the diagonals and then draw DM ⊥ AC and EN ⊥ AB.

Proof:

Consider Δ ADE and Δ BDE.

$\text{ Area of }△\mathrm{ADE}=\frac{1}{2}\times \text{ base }\times \text{ height }$

$=\frac{1}{2}\times \mathrm{AD}\times \text{ height }$

$=\frac{1}{2}\times \mathrm{AD}\times \mathrm{EN}$

$\text{ Area of }△\mathrm{BDE}=\frac{\frac{1}{2}}{2}\times \text{ base }\times \text{ height }$

$=\frac{1}{2}\times \mathrm{DB}\times \text{ height }$

$=\frac{1}{2}\times \mathrm{DB}\times \mathrm{EN}$

Taking ratios of Areas

$\frac{\text{ Area of }△\mathrm{ADE}}{\text{ Area of }△\mathrm{BDE}}=\frac{\frac{1}{\frac{1}{2}}\times \mathrm{AD}\times \mathrm{EN}}{\frac{1}{2}\times \mathrm{DB}\times \mathrm{EN}}$

$\frac{\text{ Area of }△\mathrm{ADE}}{\text{ Area of }△\mathrm{BDE}}=\frac{\mathrm{AD}}{\mathrm{DB}}$ … (1)

Now, Consider Δ AED and Δ DEC.

$\text{ Area of }△\mathrm{AED}=\frac{\frac{1}{2}}{2}\times \text{ base }\times \text{ height }$

$=\frac{1}{2}\times \mathrm{AE}\times \text{ height }$

$=\frac{1}{2}\times \mathrm{AE}\times \mathrm{DM}$

$\text{ Area of }△\mathrm{DEC}=\frac{1}{2}\times \text{ base }\times \text{ height }$

$=\frac{1}{2}\times \mathrm{EC}\times \text{ height }$

$=\frac{1}{2}\times \mathrm{EC}\times \mathrm{DM}$

Taking ratios of Areas

$\frac{\text{ Area of }△\mathrm{AED}}{\text{ Area of }△\mathrm{DEC}}=\frac{\frac{1}{2}\times \mathrm{AE}\times \mathrm{DM}}{\frac{1}{2}\times \mathrm{EC}\times \mathrm{DM}}$

$\frac{\text{ Area of }△\mathrm{AED}}{\text{ Area of }△\mathrm{DEC}}=\frac{\mathrm{AE}}{\mathrm{EC}}$ … (2)

Note that Δ BDE and DEC are on the same base DE and between the same parallels BC and DE.

So, ar Δ BDE = ar Δ DEC 

Therefore, we can write:

\frac{A D}{D B}=\frac{A E}{E C}

This completes the proof of the Basic Proportionality Theorem.

4.0Solved Examples of Basic Proportionality Theorem

Example 1: In triangle ABC, a line DE is drawn parallel to side BC, intersecting AB at D and AC at E. If AD = 3 cm, DB = 6 cm, and AE = 4 cm, find EC.

Solution:

Using the Basic Proportionality Theorem:

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

Substitute the known values:

$\frac{3}{6}=\frac{4}{EC}$

Simplifying the ratio:

$\frac{1}{2}=\frac{4}{EC}$

Cross-multiply:

EC = 8 cm

Example 2: In triangle PQR, a line ST is drawn parallel to side QR, intersecting PQ at S and PR at T. If PS = 5 cm, SQ = 10 cm, and PT = 7.5 cm, find TR.

Solution:

From the Basic Proportionality Theorem:

$\frac{PS}{SQ}=\frac{PT}{TR}$

Substitute the known values:

$\frac{5}{10}=\frac{7.5}{\mathrm{TR}}$

Simplifying the ratio:

$\frac{1}{2}=\frac{7.5}{TR}$

Cross-multiply:

TR = 15 cm

Example 3: In triangle ABC, a line DE is drawn parallel to side BC, intersecting AB at D and AC at E. If AD = 4 cm, DB = 6 cm, and AE = 5 cm, find EC.

Solution:

Given:  

AD = 4 cm  

DB = 6 cm  

AE = 5 cm  

We need to find EC.

Using the Basic Proportionality Theorem, we know:

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

Substitute the known values:

$\frac{4}{6}=\frac{5}{EC}$

Simplify the ratio:

$\frac{2}{3}=\frac{5}{EC}$

Now, cross-multiply:

2 × EC = 3 × 5

2 × EC = 15

$EC=\frac{15}{2}=7.5\mathrm{cm}$

Thus, EC = 7.5 cm.

Example 4: In triangle PQR, ST is drawn parallel to side QR, intersecting PQ at S and PR at T. If PS = 4 cm, SQ = 6 cm, and PT = 5 cm, find TR.

Solution:

Given:  

PS = 4 cm  

SQ = 6 cm  

PT = 5 cm  

We need to find TR.

According to the Basic Proportionality Theorem:

$\frac{PS}{SQ}=\frac{PT}{TR}$

Substitute the known values:

$\frac{4}{6}=\frac{5}{TR}$

Simplify the ratio:

$\frac{2}{3}=\frac{5}{TR}$

Now, cross-multiply:

$2\times TR=3\times 5$

$2\times TR=15$

$TR=\frac{15}{2}=7.5\mathrm{cm}$

Thus, TR = 7.5 cm.

Example 5: In triangle XYZ, a line DE is drawn parallel to side YZ, intersecting XY at D and XZ at E. If XD = 6 cm, DY = 9 cm, and XE = 8 cm, find EZ.

Solution:

Given:  

XD = 6 cm  

DY = 9 cm  

XE = 8 cm  

We need to find EZ.

By the Basic Proportionality Theorem:

$\frac{XD}{DY}=\frac{XE}{EZ}$

Substitute the known values:

$\frac{6}{9}=\frac{8}{EZ}$

Simplify the ratio:

$\frac{2}{3}=\frac{8}{EZ}$

Now, cross-multiply:

$2\times EZ=3\times 8$

$2\times EZ=24$

$EZ=\frac{24}{2}=12\mathrm{cm}$

Thus, EZ = 12 cm.

Example 6: In triangle DEF, a line GH is drawn parallel to side EF, intersecting DE at G and DF at H. If DG = 7 cm, GE = 14 cm, and GH = 10 cm, find EF.

Solution:

Given:  

DG = 7 cm  

GE = 14 cm  

GH = 10 cm  

We need to find EF.

By the Basic Proportionality Theorem:

$\frac{DG}{GE}=\frac{DH}{HF}$

Since GH is parallel to EF, we can apply the same proportionality for EF.

$\frac{DG}{GE}=\frac{GH}{HF}$

Substitute the known values:

$\frac{7}{14}=\frac{10}{HF}$

Simplify the ratio:

$\frac{1}{2}=\frac{10}{HF}$

Now, cross-multiply:

HF = 20 cm

Thus, EF = GH + HF = 10 + 20 = 30 cm.

Example 7: In triangle ABC, line DE is drawn parallel to BC, intersecting AB at D and AC at E. If AD = 3 cm, DB = 9 cm, and AE = 4 cm, find EC.

Solution:

Given:  

AD = 3 cm  

DB = 9 cm  

AE = 4 cm  

We need to find EC.

Using the Basic Proportionality Theorem:

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

Substitute the known values:

$\frac{3}{9}=\frac{4}{EC}$

Simplify the ratio:

$\frac{1}{3}=\frac{4}{EC}$

Now, cross-multiply:

$1\times EC=3\times 4$

$EC=12\mathrm{cm}$

Thus, EC = 12 cm.

5.0Practice Questions on Basic Proportionality Theorem

a. In triangle ABC, a line DE is drawn parallel to side BC, intersecting AB at D and AC at E. If AD = 4 cm, DB = 8 cm, and AE = 5 cm, find the length of EC.

b. In triangle XYZ, line PQ is drawn parallel to side YZ, intersecting XY at P and XZ at Q. If XP = 6 cm, PY = 9 cm, and XQ = 8 cm, find the length of QZ.

c. In triangle DEF, a line GH is drawn parallel to side EF, intersecting DE at G and DF at H. If DG = 7 cm, GE = 14 cm, and GH = 10 cm, find the length of EF.

d. In triangle PQR, a line ST is drawn parallel to side QR, intersecting PQ at S and PR at T. If PS = 3 cm, SQ = 9 cm, and PT = 5 cm, find the length of TR.

e. In triangle MNO, a line KL is drawn parallel to side NO, intersecting MN at K and MO at L. If MK = 2 cm, KN = 3 cm, and KL = 4 cm, find the length of NO.

f. In triangle ABC, line DE is parallel to BC. If AD = 5 cm, AE = 7.5 cm, and EC = 6 cm, find the length of DB.

g. In triangle PQR, a line ST is parallel to side QR and intersects PQ at S and PR at T. If PS = 4 cm, SQ = 12 cm, and PT = 6 cm, find the length of TR.

h. In triangle XYZ, line MN is parallel to side YZ and intersects XY at M and XZ at N. If XM = 3 cm, MY = 6 cm, and XN = 4.5 cm, find the length of NZ.

i. In triangle ABC, line DE is parallel to side BC and intersects AB at D and AC at E. If AD = 4 cm, DB = 6 cm, and EC = 8 cm, find the length of AE.

j. In triangle DEF, a line GH is drawn parallel to side EF, intersecting DE at G and DF at H. If DG = 10 cm, GE = 5 cm, and HF = 6 cm, find the length of DH.

Solutions:

a. EC = 10 cm 

b. QZ = 12 cm

c. EF = 21 cm  

d. TR = 15 cm  

e. NO = 10 cm  

f. DB = 4 cm  

g. TR = 18 cm  

h. NZ = 9 cm  

i. AE = 5.33 cm  

j. DH = 12 cm

6.0Sample Question on Basic Proportionality Theorem

1. What is the Basic Proportionality Theorem? 

Ans: The Basic Proportionality Theorem states that if a line is drawn parallel to one side of a triangle and intersects the other two sides, it divides those two sides in the same ratio. 

For example, in a triangle ABC, if DE is drawn parallel to side BC and intersects AB at D and AC at E, then:

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