Question 2
In the following figure, D and E are two points on BC such that BD = DE= EC. Show
that area (ABD) = area (ADE) = area (AEC).

[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 the division so obtained to the opposite vertex of BC, you can divide $Δ A B C$ into triangles of equal areas.]

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Solution

Let us draw a line segment $A M ⊥ B C$ .
We know that,
Area of a triangle $= \frac{1}{2} \times B a s e \times A l t i t u d e$
$A r e a (Δ A D E) = \frac{1}{2} \times D E \times A M$
$A r e a (Δ A B D) = \frac{1}{2} \times B D \times A M$
$A r e a (Δ A E C) = \frac{1}{2} \times E C \times A M$
It is given that, DE= BD = EC
$\Rightarrow \frac{1}{2} \times D E \times A M = \frac{1}{2} \times B D \times A M = \frac{1}{2} \times E C \times A M$
$∴ A r e a (Δ A D E) = A r e a (Δ A B D) = A r e a (Δ A E C)$

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Similar questions

In the following figure, D and E are two points on BC such that BD = DE = EC. Show that ar (ABD) = ar (ADE) = ar (AEC).

Can you 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 ΔABC into n triangles of equal areas.]

Q. In Fig,

D

and

E

are two points on

B C

such that

B D = D E = E C

. Show that

a r (A B D) = a r (A D E) = a r (A E C)

.
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?

D and E are points on the sides AB and AC, respectively, of $△$ ABC such that DE || BC and DE divides $△$ ABC into two parts that are equal in area. Find $\frac{B D}{A B}$ .