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 acute angle with BC and X lies on the opposite side of A with respect to BC. The points B_{1}, B_{2}, ...., B_{7} are located at equal distances on BX, B_{3} is joined to C and then a line segment B_{6}C' is drawn parallel to B_{3}C where C' lies on BC produced. Finally, line segment A'C' is drawn parallel to AC.

#### Options

True

False

#### Solution

This statement is **False**.

**Explanation:**

Let us try to construct the figure as given in the question.

**Steps of construction:**

**1.** Draw a line segment BC.

**2.** With B and C as centres, draw two arcs of suitable radius intersecting each other at A.

**3.** Join BA and CA and we get the required triangle ∆ABC.

**4.** Draw a ray BX from B downwards to make an acute angle ∠CBX.

**5.** Now, mark seven points B_{1}, B_{2}, B_{3} …B_{7} on BX, such that 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 draw a line B_{7}C’ || B_{3}C from B_{7} such that it intersects the extended line segment BC at C’.

**7.** Draw C’A’ ||CA in such a way that it intersects 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.

We have,

Segment B_{6}C’ || B_{3}C. But it is clear in our construction that it is never possible that segment B_{6}C’ || B_{3}C since 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.