To construct a triangle similar to a given \(\triangle ABC\) with its sides \(\dfrac{7}{3}\) of the corresponding sides of \(\triangle ABC\), draw a ray \(BX\) making an acute angle with \(BC\) and with \(X\) on the opposite side of \(A\) w.r.t. \(BC\). Locate points \(B_1,B_2,\dots,B_7\) at equal distances on \(BX\). Then join \(B_3\) to \(C\) and draw \(B_6C'\) \(\parallel\) \(B_3C\) meeting \(BC\) produced at \(C'\). Finally, draw \(A'C'\) \(\parallel\) \(AC\).
Step 1: We want a triangle similar to \(\triangle ABC\) but larger. The scale factor is \(\dfrac{7}{3}\), which means each side of the new triangle should be \(\tfrac{7}{3}\) times the corresponding side of \(\triangle ABC\).
Step 2: To enlarge a triangle from vertex \(B\), we first draw a ray \(BX\) making an acute angle with \(BC\). On this ray, we mark 7 equal divisions: \(B_1, B_2, B_3, ..., B_7\).
Step 3: Why 7 points? Because the numerator of the scale factor is 7. Why do we also consider 3? Because the denominator is 3. To enlarge, we need to use the last point \(B_7\) and also the 3rd point \(B_3\).
Step 4: The correct method is: – Join \(B_7\) to \(C\). – Through \(B_3\), draw a line parallel to \(B_7C\). This line will cut the extended line of \(BC\) at a new point, call it \(C'\).
Step 5: Now, through \(C'\), draw a line parallel to \(AC\). This line meets the extension of \(AB\) at a point \(A'\). Thus, \(\triangle A'BC'\) is the required triangle similar to \(\triangle ABC\) with sides in the ratio \(\dfrac{7}{3}\).
Step 6: In the question, they mistakenly joined \(B_3\) to \(C\) first, instead of \(B_7\) to \(C\). This mixes up the roles of \(B_3\) and \(B_7\). As a result, the construction given is wrong.
Final Check: Since the steps in the question are not the correct method for enlargement, the statement is False.