Let ∆PQR and ∆ABC are similar triangles, then its scale factor between the corresponding sides is:

Steps of construction

1. Draw a line BC=5 cm.

2. Construct OQ the perpendicular bisector of line segment BC meeting BC at P’.

3. Taking B and C as centers draw two arcs of equal radius 6 cm intersecting each other at A.

4. Join BA and CA So, ∆ABC is the required isosceles triangle.

5. Draw any ray BX making an acute ∠CBX.

6. Locate 4 points B₁,B₂,B₃ and B₄ on BX such that BB₁=B₁B₂=B₂B₃=B₃B₄

7. Join B₃C and from B₄ draw a line B₄R||B₃C intersecting the extended line segment BC at R.

8. Draw RP||CA meeting BA produced at P.

Then, ∆PBR is the required triangles.

By construction,

B₄R││B₃C

∴ BC/CR = 3/1

Now,

∴

Now,

⇒

⇒

⇒ BR/BC = 4/3

Also,

RP||CA

∴ ΔABC ~ ΔPBR

and

Hence, the new triangle is similar to the given triangle whose sides are 4/3 times of the corresponding sides of the isosceles ∆ABC.