To construct a triangle similar to a given $\triangle \mathrm{ABC}$ with its sides $\frac{7}{3}$ of the corresponding sides of $\triangle \mathrm{ABC}$,
draw a ray $B X$ making acute angle with $B C$ and $X$ lies on the opposite side of $A$ with respect of $B C$. The points $B_{1}$,
$B_{2}, \ldots, B_{7}$ are located at equal distances on $B X, B_{3}$ is joined to $C$ and then a line segment $B_{6} C^{\prime}$ is drawn parallel to
$B_{3} C$, where $C^{\prime}$ lines on $B C$ produced. Finally line segment $A^{\prime} C^{\prime}$ is drawn parallel to $A C$.
False
Steps of construction
- Draw a line segment BC with suitable length.
- Taking B and C as centres draw two arcs of suitable radii intersecting each other at A
- Join BA and CA ΔABC is the required triangle.
- From B draw any ray BX downwards making an acute angle CBX.
- Locate seven points B1, B2, …, B7 on SX, such that BB1 = B1B2 = B2B3 = B3B4 = B4B5 = B5B6 = B6B7.
- Join B3C and from B7 draw a line B7C’|| B3C intersecting the extended line segment BC at C’.
- From point C’ draw C’A’|| CA intersecting the extended line segment BA at A’.Then, ΔA’BC’ is the
required triangle whose sides are $\frac{7}{3}$ of the corresponding sides of $\triangle \mathrm{ABC}$.
Given that, segment $B_{6} C^{\prime}$ is drawn parallel to $B_{3} C$. But from our construction is never possible that segment $B_{6} C^{\prime}$
is parallel to $B_{3} C$ because the similar triangle $A^{\prime} B C^{\prime}$ has its sides $\frac{7}{3}$ of the corresponding sides of triangle
$\mathrm{ABC}$. So, $\mathrm{B}_{7} \mathrm{C}^{\prime}$ is parallel to $\mathrm{B}_{3} \mathrm{C}$.
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