##### To construct a triangle similar to a given ∆ ABC with its sides 7/3 of the corresponding sides of ∆ABC, draw a ray BX making an acute angle with BC and X lies on the opposite side of A with respect of BC. The points B1, B2,...........B1 are located at equal distances on BX, B3 is joined to C and then a where C’ lines on BC produced. Finally line segment A’C’ is drawn parallel to AC.

False

Steps of construction

1. Draw a line segment BC.

2. B and C as centers draw two arcs of suitable radius intersecting each other at A.

3. Join BA and CA. ∆ABC is required triangle.

4. From B draw any ray BX downwards making an acute angle CBX.

5. Marked seven points B1,B2,B3,............B7 on BX (BB1 = B1B2 = B2B3 = B3B4 = B4B5 = B5B6 = B6B7).

6. Join B3C and from B7 draw a line B7C’B3C intersecting the extended line segment BC at C’.

7. Draw C’ A’CA intersecting the extended line segment BA at A’.

Then, ∆A’BC’ is the required triangle whose sides are 7/3 of the corresponding sides of ∆ABC.

Given that,

segment B6C’ B3C. But since our construction is never possible that segment B6C’ B3C because the similar triangle A’BC’ has its sides 7/3 of the corresponding sides of triangle ABC.

So, B7C’ is parallel to B3C.

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