Draw a line segment AB of length 7cm. Using ruler and compasses, find a point P on AB such that .

Steps of construction:

1. Draw a line segment AB = 7 cm and Below AB, Draw an acute angle ∠BAQ.

2. Mark five points A₁, A₂, …, A₅ on AX such that AA₁ = A₁A₂ = … = A₄A₅.

3. Join AA₅.

Now, we have to draw a line parallel to AA₅ from point A₃ and for that

4. With A₅ as center and with any measure, draw an arc interesting lines AQ and AA₅ at X and Y respectively.

5. With A₃ as center and with same measure as in previous step draw an arc which intersect AQ at Z.

6. Taking Z as center, and with the measure XY, draw another arc which intersect the previous arc at W.

7. Make a line segment from A₃ and passing through W, which intersects AB at P.

Verification:

Here, we have AA₃ || AA₅

So, By Basic Proportionality theorem in ΔABX, we have

i.e. AP = 3x and BP = 2x

⇒ AB = AP + BP = 5x

and