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.