Construction Procedure:

1. Draw a line segment AB which measures 4 cm, i.e., AB = 4 cm.

2. Take the point A as centre, and draw an arc of radius 5 cm.

3. Similarly, take the point B as its centre, and draw an arc of radius 6 cm.

4. The arcs drawn will intersect each other at point C.

5. Now, we obtained AC = 5 cm and BC = 6 cm and therefore ΔABC is the required triangle.

6. Draw a ray AX which makes an acute angle with the line segment AB on the opposite side of vertex C.

7. Locate 3 points such as A1, A2, A3 (as 3 is greater between 2 and 3) on line AX such that it becomes AA1= A1A2 = A2A3.

8. Join the point BA3 and draw a line through A2which is parallel to the line BA3 that intersect AB at point B’.

9. Through the point B’, draw a line parallel to the line BC that intersect the line AC at C’.

10. Therefore, ΔAB’C’ is the required triangle.

Justification:

The construction of the given problem can be justified by proving that

AB’ = (2/3)AB

B’C’ = (2/3)BC

AC’= (2/3)AC

From the construction, we get B’C’ || BC

∴ ∠AB’C’ = ∠ABC (Corresponding angles)

In ΔAB’C’ and ΔABC,

∠ABC = ∠AB’C (Proved above)

∠BAC = ∠B’AC’ (Common)

∴ ΔAB’C’ ∼ ΔABC (From AA similarity criterion)

Therefore, AB’/AB = B’C’/BC= AC’/AC …. (1)

In ΔAAB’ and ΔAAB,

∠A₂AB’ =∠A₃AB (Common)

From the corresponding angles, we get,

∠AA₂B’ =∠AA₃B

Therefore, from the AA similarity criterion, we obtain

ΔAA₂B’ and AA₃B

So, AB’/AB = AA₂/AA₃

Therefore, AB’/AB = 2/3 ……. (2)

From the equations (1) and (2), we get

AB’/AB=B’C’/BC = AC’/ AC = 2/3

This can be written as

AB’ = (2/3)AB

B’C’ = (2/3)BC

AC’= (2/3)AC

Hence, justified.