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 B_{1}, B_{2},...........B_{1} are located at equal distances on BX, B_{3} 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 B_{1},B_{2},B_{3},............B_{7} on BX (BB_{1} = B_{1}B_{2} = B_{2}B_{3} = B_{3}B_{4} = B_{4}B_{5} = B_{5}B_{6} = B_{6}B_{7}).

6. Join B_{3}C and from B_{7} draw a line B_{7}C’B_{3}C 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 B_{6}C’ B_{3}C. But since our construction is never possible that segment B_{6}C’ B_{3}C because the similar triangle A’BC’ has its sides 7/3 of the corresponding sides of triangle ABC.

So, B_{7}C’ is parallel to B_{3}C.

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