#### Question

Divide a line segment of length 14 cm internally in the ratio 2 : 5. Also, justify your construction.

#### Solution

Given that

Determine a point which divides a line segment of length 14cm internally in the ratio of 2:5.

We follow the following steps to construct the given

Step of construction

Step: I-First of all we draw a line segment AB = 14cm.

Step: II- We draw a ray AX making an acute angle ∠BAX = 60° with AB.

Step: III- Draw a ray BY parallel to *AX* by making an acute angle ∠ABY = ∠BAX.

Step IV- Mark of two points A_{1}, A_{2} on AX and three points B_{1}, B_{2}, B_{3}, B_{4}, B_{5} on BY in such a way that AAA1 = A1A2 = B1B2 = B_{2}B_{3} = B_{3}B_{4} = B_{4}B_{5}.

Step: V- Joins A_{2}B_{3} and this line intersects AB at a point P.

Thus, *P* is the point dividing AB internally in the ratio of 2:5

Justification:-

In ΔAA_{2}P and ΔBB_{5}P we have

∠A_{2}AP = ∠PBB_{5} [∠ABY = ∠BAX]

And ∠APA_{2} = ∠BPB_{5} [Vertically opposite angle]

So, AA similarity criterion, we have

ΔAA_{2}P ≈ ΔBB_{5}P

`(A A_2)/(BB_5)=(AP)(BP)`

`(AP)/(BP)=2/5`