Question

Mention the proper order of identifying a point D on BC such that $\frac{Ar\left(△ABD\right)}{Ar\left(△ADC\right)}=\frac{2}{3}$?
1) Join $B_{5}C$ and draw a line parallel to $B_{5}C$ from $B_2$.
2) Identify the ratio in which the point D divides BC using the relation given.
3) Mark 5 points B₁, B₂, B₃, B₄ and B₅ on BX such that they are equidistant.
4) Construct a ray BX which makes an acute angle with line segment BC.
5) The point of intersection of the parallel line from B₂ with BC is the point D.

• A. 2, 4, 3, 1, 5
• B. 1, 2, 4, 3, 5
• C. 4, 3, 2, 1, 5
• D. 3, 5, 2, 4, 1

Answer 1

The correct option is A

2, 4, 3, 1, 5

$\frac{Ar\left(ABD\right)}{Ar\left(ADC\right)}=\frac{\left(\frac{1}{2}\right)\left(BasedBD\right)\left(HightAE\right)}{\left(\frac{1}{2}\right)\left(BaseDC\right)\left(HeightAE\right)}=\frac{BD}{DC}=\frac{2}{3}$ i.e. D divides BC in the ratio 2:3. Our next obvious task, is therefore to find this point D. So, draw a raywhich makes an acute angle with side BC. Now perform the constructions required to mark D on BC such that BD:DC = 2:3.


The steps of construction:
First you need to construct the $△$ABC. Now that you have constructed the $△$ABC we can divide the line BC such that BD:DC = 2:3.
a) Let us start by constructing a ray BX which makes an acute angle with line segment BC.
b) Now mark $B_1,B_2,B_3,B_4$ and $B_5$ on ray BX such that $B_1{B}_2=B_2{B}_3=B_3{B}_4=B_4{B}_5$. (This is done in order to divide $B{B}_2$ and $B_2{B}_5$ in the ratio 2:3)
c) Now join $B_{5}C$ and draw $B_{2}D$ parallel to $B_{5}C$ such that the point D lies on the line segment BC. By Basic Proportionality Theorem, $\frac{BD}{DC}=\frac{B{B}_2}{B_2{B}_5}=\frac{2}{3}$.
The ratio of the area of the resulting triangles $△ABD$ and $△ADC$ will be $\frac{2}{3}$.
So, the correct order is 2, 4, 3, 1, 5.