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Steps of Construction:

Step I: BC = 5 cm is drawn.

Step II: Draw perpendicular bisector of BC, Let the bisector intersect BC at O.

Taking point D as center, and 4 cm radius, draw an arc interesting bisector at point A.

Join AB and AC.

Step III: A ray BX is drawn making an acute angle with BC opposite to vertex A.

Step IV: 3 points B1 , B2 and B3 is marked BX.

Step V: B2 is joined with C to form B2C as 2 point is smaller. B3C' is drawn parallel to B2C as 3 point is greater.

Step VI: C'A' is drawn parallel to CA.

Thus, A'BC' is the required triangle formed.

Justification:

Since the scale factor is ,

We need to prove,

By construction,

… (1)

Also, A1C1 is parallel to AC.

So, this will make same angle with BC.

A1C1B = ACB …. (2)

Now,

In ΔA1BC1 and ΔABC

B = B (common)

A1C1B = ACB (from 2)

ΔA1BC1 ΔABC

Since corresponding sides of similar triangles are in same ratio.

From (1)

Hence construction is justified.

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