False

Support:

Allow us to attempt to build the figure as given in the inquiry.

Steps of development,

1. Define a boundary section \[BC.\]
2. With \[B\text{ }and\text{ }C\]as focuses, draw two circular segments of appropriate sweep crossing each other at\[A\].
3. Join \[BA\text{ }and\text{ }CA\]and we get the necessary triangle \[\vartriangle \mathbf{ABC}\]
4. Draw a beam \[BX\]from \[B\]downwards to make an intense point \[\angle CBX.\]
5. Presently, mark seven focuses\[{{B}_{1}},\text{ }{{B}_{2}},\text{ }{{B}_{3}}~\ldots {{B}_{7}}~on\text{ }BX\]

   to such an extent that\[B{{B}_{1}}~=\text{ }{{B}_{1}}{{B}_{2}}~=\text{ }{{B}_{2}}{{B}_{3}}~=\text{ }{{B}_{3}}{{B}_{4}}~=\text{ }{{B}_{4}}{{B}_{5}}~=\text{ }{{B}_{5}}{{B}_{6}}~=\text{ }{{B}_{6}}{{B}_{7}}.\]

6. Join\[{{B}_{3}}C\]

   and define a boundary \[{{B}_{7}}C||\text{ }{{B}_{3}}C\] from  \[{{B}_{7}}\] with the end goal that it converges the lengthy line fragment \[BC\text{ }at\text{ }C’.\]

7. Attract \[C’A’\text{ }||CA\] such a way that it meets the drawn out line section \[BA\text{ }at\text{ }A’.\]

Then, at that point, \[\vartriangle \]\[~A’BC’\] is the necessary triangle whose sides are \[7/3\] of the comparing sides of \[\vartriangle \mathbf{ABC}\].

As per the inquiry,

We have,

Portion\[{{B}_{6}}C\text{ }||\text{ }{{B}_{3}}C\] In any case, it is clear in our development that it is never conceivable that fragment\[{{B}_{6}}C||{{B}_{3}}C\] since the comparative triangle \[A’BC’\] has its sides \[7/3\]of the relating sides of triangle \[ABC.\]

Along these lines,

\[{{B}_{7}}C\]

is corresponding to

\[{{B}_{3}}C.\]