Construct a triangle similar to a given triangle \(ABC\) with its sides equal to \(\frac{6}{5}\) of the corresponding sides of the triangle \(ABC\) (scale factor \(\frac{6}{5} > 1\))
Construction:
Step 1:
Step 2:
Step 3:
Step 4:
Step 5:
Then, \(A'BC'\) is the required triangle, each of whose side is six - fifth of the corresponding sides of \(\triangle ABC\).
Answer variants:
Join \(Q_5\) (the fifth point, \(5\) being smaller of \(5\) and \(6\) in \(\frac{6}{5}\)) to \(C\) and draw a line through \(Q_6\) parallel to \(Q_5C\), intersecting the extended line segment \(BC\) at \(C'\).
Draw a line through \(C'\) parallel to the line \(AC\) to intersect \(AB\) at \(A'\).
Construct a triangle \(ABC\) with any measurement.
Draw a ray \(BX\), making an acute angle with \(BC\) on the side opposite to vertex \(A\).
Locate \(5\) points (the greater of \(6\) and \(5\) in \(\frac{6}{5}\)). \(Q_1\), \(Q_2\), \(Q_3\), \(Q_4\), \(Q_5\), and \(Q_6\) on \(BX\) so that \(BQ_1 = Q_1Q_2 = Q_2Q_3 = Q_3Q_4 = Q_4Q_5 = Q_5Q_6\).