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In Fig. 9.23, BD || CA, E is mid-point of CA and BD = 1/2 CA. Prove that ar (ABC) = 2ar (DBC).

Solution:

Given, BD || CA

E is the midpoint of CA

BD = 1/2 CA

We have to prove that ar(ABC) = 2 ar(DBC)

Join DE

Now, BD = CE

BD || CE

We know that the opposite sides of a parallelogram are parallel and congruent.

So, BCED is a parallelogram

We know that the area of triangles on the same base and between the same parallel lines are equal.

Triangles DBC and EBC are on the same base BC and between the same parallel lines AD and BC.

So, ar(DBC) = ar(EBC) ------- (1)

Considering triangle ABC,

BE is the median of the triangle ABC

We know that the median of a triangle divides it into two triangles of equal areas.

ar(EBC) = 1/2 ar(ABC)

So, ar(ABC) = 2 ar(EBC)

From (1),

ar(ABC) = 2 ar(DBC)

Therefore, it is proved that ar(ABC) = 2 ar(DBC)

✦ Try This: ABCD is a trapezium in which AB || DC, AB = 16cm and DC = 24cm. If E and F are respectively the midpoints of AD and BC, prove that ar (ABFE) = 9/11 ar (EFCD).

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

NCERT Exemplar Class 9 Maths Exercise 9.4 Sample Problem 3

In Fig. 9.23, BD || CA, E is mid-point of CA and BD = 1/2 CA. Prove that ar (ABC) = 2ar (DBC).

Summary:

In Fig. 9.23, BD || CA, E is mid-point of CA and BD = 1/2 CA. It is proven that ar (ABC) = 2ar (DBC) as we know that the area of the triangles on the same base and between the same parallel lines are equal

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