Basic Proportionality Theorem (BPT)

Basic Proportionality Theorem: Thales' theorem is one of the most fundamental theorems in geometry that relates the parts of the length of sides of triangles. The other name of the Thales theorem is the Basic Proportionality Theorem or BPT.

BPT states that if a line is parallel to a side of a triangle that intersects the other sides into two distinct points, then the line divides those sides in proportion.

Basic Proportionality Theorem or Thales Theorem Statement

Thales's Theorem or Basic Proportionality Theorem (BPT) states that if a line is drawn parallel to one side of a triangle intersecting the other two sides in distinct points, then the other two sides are divided in the same ratio.

Basic Proportionality Theorem Proof

Let's prove the Basic Proportionality Theorem.

Given: Let us suppose we have a triangle ABC, if we draw a line LM parallel to side BC.

To Proof: \frac{AL}{LB}=\frac{AM}{MC}               

Construction

To prove the required result, construct the following lines in the given figure:

• Construct LY⊥AM.
• Construct MX⊥AL.
• Join points L with C and M with B with a line segment.

Basic Proportionality Theorem Proof

Since, area of triangle= (\frac{1}{2}\times base\times height)

Area of \triangle ALM=(\frac{1}{2}\times AL\times MX)

Area of \triangle LBM=(\frac{1}{2}\times LB\times MX)

Area of \triangle ALM=(\frac{1}{2}\times AM\times LY)

Area of \triangle LMC=(\frac{1}{2}\times MC\times LY)

Ratio of area of \triangle ALM and \triangle LBM:

\frac{area(\triangle ALM)}{area(\triangle LBM)}=\frac{(\frac{1}{2}\times AL\times MX)}{(\frac{1}{2}\times LB\times MX)}=\frac{AL}{LB}. . .(1)

Ratio of area of \triangle ALM and  LMC:

\frac{area(\triangle ALM)}{area(\triangle LMC)}=\frac{(\frac{1}{2}\times AM\times LY)}{(\frac{1}{2}\times MC\times LY)}=\frac{AM}{MC}. . .(2)

According to the property of triangles, the triangles on the same base and between same parallel lines have equal areas.

Therefore, \triangle LBM and \triangle LMC have equal areas. 

i.e., area of \triangle  LBM = area of \triangle LMC. . .(3)

From equations (1),(2), and (3) we can conclude:

\bold{\frac{AL}{LB}=\frac{AM}{MC}}  [Hence Proved]

Corollary of Thales Theorem

Mid Point theorem states that, "If the line is drawn from one of the midpoints of the side of a triangle parallel to the second, then it always intersects the third side at the midpoint as well."

Related Articles:

• Mid Point theorem
• Figures on the Same Base and between the Same Parallels
• Criteria for Similarity of Triangles

Converse of Basic Proportionality Theorem (BPT)

The Converse of the Basic Proportionality Theorem (BPT) or Thales's Theorem states that if a line intersects two sides of a triangle and divides the sides proportionally, then the line is parallel to the third side of the triangle.

Proof

Let's prove the converse of the Basic Proportionality Theorem or converse of Thales Theorem is given as follows:

Given: Let us consider a triangle ABC with D and E two points on the side AB and AC such that AD/BD = AE/CE . . .(i).

To Prove: DE∥BC

Construction

To prove the converse of BPT, draw a line DE'∥BC.

Proof

Now, as DE'∥BC in triangle ABC, 

Thus, by using the Basic proportionality theorem,

AD/BD = AE'/CE' . . .(ii)

Using equations (i) and (ii), we get

AE/CE = AE'/CE'

Add 1 to both sides of the equation,

AE/CE + 1 = AE'/CE' + 1

⇒ (AE+CE)/CE = (AE'+CE')/CE'

⇒ AC/CE = AC/CE'

⇒ CE = CE'

This can only be possible if E and E' are coincident.

Thus, E and E' are the same points. 

As, DE∥BC 

So, DE∥BC [Hence Proved]

Solved Examples on Basic Proportionality Theorem

Let's solve some questions on the Basic Proportionality Theorem.

Example 1: In ΔPQR, X and Y are points on the sides PQ and PR respectively such that XY|| QR. If PX/XQ = 2/5 and PR = 10 cm find PY. 

Solution:

In the given triangle PQR, X and Y are points on the sides PQ and PR respectively such that XY || QR.

By the Basic Proportionality Theorem, we have:

PX/XQ = PY/YR

⇒ PX/XQ = PY/(PR-PY)

Given that PX/XQ = 2/5, we can substitute for PX/XQ:

2/5 = PY/(PR-PY)

⇒ 2/5 = PY/(10-PY)

⇒ 2/5 = PY/(10-PY)

⇒ 2(10 - PY) = 5PY

⇒ 20 - 2PY = 5PY

⇒ 7PY = 20 

⇒ PY = 20 /7

Hence, the length of PY is 20/7 cm.

Example 2: In ΔPQR, X and Y are points on the sides PQ and PR respectively such that XY|| QR. If PX = x − 7 , XQ = x + 3 , PY = 2x and YR = 2x −1 , find the value of x.

Solution:

In the given triangle PQR, X and Y are points on the sides PQ and PR respectively such that XY || QR.

By the Basic Proportionality Theorem, we have:

PX/XQ = PY/YR

⇒ (x - 7)/(x + 3) = 2x/(2x - 1)

⇒ (x - 7)(2x - 1) = 2x(x + 3)

⇒ 2x² - x - 14x +7 = 2x² + 6x

⇒ 2x² - 15x +7 = 2x² + 6x

Subtract 2x² from both sides:

- 15 x + 7 = 6x

Bring terms together:

7 = 6x + 15

7 = 21 x

x = 7/21

x = 1/3

Example 3: In an ∆ABC, sides AB and AC  are intersected by a line at D and E respectively, which is parallel to side BC. Prove that AD/AB = AE/AC.

Solution:

Given: DE || BC. So, AD/DB = AE/EC

or By interchanging the ratios as => DB/AD = EC/AE

Now, add 1 on both sides 

(DB/AD) + 1 = (EC/AE) + 1

⇒ (DB + AD)/AD = (EC + AE)/AE

⇒ AB/AD = AC/ AE

If we interchange the ratios again, we get

AD/AB = AE/AC [Hence Proved]

Example 4: In triangle ABC, where DE is a line drawn from the midpoint of AB and ends midpoint of AC at E. AD/DB = AE/EC and ∠ADE = ∠ACB. Then prove ABC is an isosceles triangle.

Solution:

Given: AD/DB = AE/EC

By the converse of the basic proportionality theorem, we get => DE || BC

According to question; ∠ADE = ∠ACB

Hence, ∠ABC = ∠ACB

The side opposite to equal angles is also equal to AB = AC

Hence, ABC is an isosceles triangle.

Practice Problems on Basic Proportionality Theorem

Question 1: In triangle ABC, a line DE is drawn parallel to BC and intersects AB at D and AC at E. If AD = 3 cm, DB = 2 cm, and AE = 4.5 cm, find the length of EC.

Question 2: In triangle XYZ, line LM is parallel to YZ, intersecting XY at L and XZ at M. If XL = 5cm, LY = 10 cm, and XM = 8 cm, find the length of MZ.

Question 3: In triangle DEF, line GH is parallel to EF and intersects DE at G and DF at H. If DG = 6 cm, GE = 9 cm, and DH = 8 cm, find the length of HF.