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# In Fig. 9.30, D and E are two points on BC such that BD = DE = EC. Show that ar (ABD) = ar (ADE) = ar (AEC). Can you now answer the question that you have left in the ‘Introduction’ of this chapter, whether the field of Budhia has been actually divided into three parts of equal area?

[Remark: Note that by taking BD = DE = EC, the triangle ABC is divided into three triangles ABD, ADE and AEC of equal areas. In the same way, by dividing BC into n equal parts and joining the points of division so obtained to the opposite vertex of BC, you can divide ∆ABC into n triangles of equal areas.]

**Solution:**

Let's draw an altitude that will help us to find areas of all three triangles.

Let us draw AL ⊥ BC.

We know that,

Area of a triangle = 1/2 × Base × Altitude

Area (ΔADE) = 1/2 × DE × AL

Area (ΔABD) = 1/2 × BD × AL

Area (ΔAEC) = 1/2 × EC × AL

DE = BD = EC (Given)

Thus, 1/2 × DE × AL = 1/2 × BD × AL = 1/2 × EC × AL

Therefore, ar (ΔADE) = ar (ΔABD) = ar (ΔAEC)

Yes, we can now say that the field of Budhia has been actually divided into three parts of equal area.

**☛ Check: **NCERT Solutions for Class 9 Maths Chapter 9

**Video Solution:**

## In Fig. 9.30, D and E are two points on BC such that BD = DE = EC. Show that ar (ABD) = ar (ADE) = ar (AEC). Can you now answer the question that you have left in the ‘Introduction’ of this chapter, whether the field of Budhia has been actually divided into three parts of equal area? [Remark: Note that by taking BD = DE = EC, the triangle ABC is divided into three triangles ABD, ADE and AEC of equal areas. In the same way, by dividing BC into n equal parts and joining the points of division so obtained to the opposite vertex of BC, you can divide ∆ABC into n triangles of equal areas.]

Maths NCERT Solutions Class 9 Chapter 9 Exercise 9.4 Question 2

**Summary:**

If D and E are two points on BC in the given figure such that BD = DE = EC, then ar (ABD) = ar (ADE) = ar (AEC). Yes, we can now say that the field of Budhia has been actually divided into three parts of equal area.

**☛ Related Questions:**

- In Fig. 9.31, ABCD, DCFE and ABFE are parallelograms. Show that ar (ADE) = ar (BCF).
- In Fig. 9.32, ABCD is a parallelogram and BC is produced to a point Q such that AD = CQ. If AQ intersect DC at P, show that ar (BPC) = ar (DPQ).[Hint: Join AC.]
- In Fig.9.33, ABC and BDE are two equilateral triangles such that D is the mid-point of BC. If AE intersects BC at F, show thati) ar (BDE) =1/4 ar (ABC)ii) ar (BDE) = 1/2 ar (BAE)iii) ar (ABC) = 2 ar (BEC)iv) ar (BFE) = ar (AFD)v) ar (BFE) = 2 ar (FED)vi) ar (FED) = 1/8 ar (AFC)[Hint : Join EC and AD. Show that BE || AC and DE || AB, etc.]
- Diagonals AC and BD of a quadrilateral ABCD intersect each other at P. Show that ar (ΔAPB) × ar (ΔCPD) = ar (ΔAPD) × ar (ΔBPC)[Hint: From A and C, draw perpendiculars to BD.]

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