Basic Proportionality Theorem

Here we will learn how to prove the basic proportionality theorem with diagram.

A line drawn parallel to one side of a triangle divides the other two sides proportionally.

Given: In ∆XYZ, P and Q are points on XY and XZ respectively, such that PQ ∥ YZ.

Basic Proportionality Theorem

To prove: \(\frac{XP}{PY}\) = \(\frac{XQ}{QZ}\).

Proof:

            Statement

            Reason

1. In ∆XYZ and ∆XPQ,

(i) ∠YXZ = ∠PXQ

(ii) ∠XYZ = ∠XPQ

1.

(i) Common angle

(ii) Corresponding angles

2. ∆XYZ ∼ ∆XPQ

2. AA criterion of similarity.

3. \(\frac{XY}{XP}\) = \(\frac{XZ}{XQ}\)

3. Corresponding sides of similar triangles are proportional.

4. \(\frac{XY}{XP}\) – 1 = \(\frac{XZ}{XQ}\) – 1

⟹ \(\frac{XY - XP}{XP}\) = \(\frac{XZ - XQ}{XQ}\)

⟹ \(\frac{PY}{XP}\) = \(\frac{QZ}{XQ}\)

4. By subtracting 1 from both sides of statement 3.

5. \(\frac{XP}{PY}\) = \(\frac{XQ}{QZ}\)

5. Taking reciprocals of both sides in statement 4.


Solved examples using basic proportionality theorem:

1. If in a ∆XYZ, P and Q are two points on XY and XZ respectively such that XP = 4 cm, PY = 3 cm, XQ = = 6 cm, QZ = 4.5 cm and ∠XPQ = 40° then find ∠XYZ.

Problems on Basic Proportionality Theorem

Solution:

Here, \(\frac{XP}{PY}\) = \(\frac{4 cm}{3 cm}\) = \(\frac{4}{3}\), and

\(\frac{XQ}{QZ}\) = \(\frac{6 cm}{4.5 cm}\) = \(\frac{4}{3}\)

Therefore, \(\frac{XP}{PY}\) = \(\frac{XQ}{QZ}\)

⟹ PQ ∥ YZ

Therefore,  ∠XYZ = ∠XPQ = 40°.


2. In the given figure, if XP = 6 cm, YP = 2 cm, XQ = 7.5 cm, find QZ.

Numerical Problems on Basic Proportionality Theorem

Solution:

By basic proportionality theorem,

\(\frac{XP}{PY}\) = \(\frac{XQ}{QZ}\)

⟹ \(\frac{6 cm}{2 cm}\) = \(\frac{7.5 cm}{QZ}\)

⟹ QZ = \(\frac{7.5 cm × 2}{6}\)

⟹ QZ = 2.5 cm.


3. At a certain time of the day, a man, 6 feet tall, casts his shadow 8 feet long. Find the length of the shadow cast by a building 45 feet high, at the same time.

Solution:

Let the length of the shadow of the building be x.

Basic Proportionality Theorem Problem

As the source of light is the sun, XZ ∥ PQ and, hence ∆YXZ ∼ ∆YPQ.

Therefore, \(\frac{\textrm{Height of the Man}}{\textrm{Height of the Building}}\) = \(\frac{\textrm{Length of Shadow Cast by the Man}}{\textrm{Length of Shadow Cast by the Building}}\)

⟹ \(\frac{6 ft}{45 ft}\) = \(\frac{8 ft}{x}\)

⟹ x = 60 feet.







9th Grade Math

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