Mention the proper order of identifying a point D on BC such that Ar(â–³ABD)Ar(â–³ADC)=23?
1) Join B5C and draw a line parallel to B5C from B2.
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.
2, 4, 3, 1, 5
Ar(ABD)Ar(ADC)=(12)(BasedBD)(HightAE)(12)(BaseDC)(HeightAE)=BDDC=23 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 B1,B2,B3,B4 and B5 on ray BX such that B1B2=B2B3=B3B4=B4B5. (This is done in order to divide BB2 and B2B5 in the ratio 2:3)
c) Now join B5C and draw B2D parallel to B5C such that the point D lies on the line segment BC. By Basic Proportionality Theorem, BDDC=BB2B2B5=23.
The ratio of the area of the resulting triangles △ABD and △ADC will be 23.
So, the correct order is 2, 4, 3, 1, 5.