Step-by-Step Construction (reduction by \(\dfrac{5}{7}\))
- Draw the given triangle.
- First, draw a line segment \(AB = 5\,\text{cm}\).
- At point \(B\), use a protractor to make an angle of \(60^\circ\).
- From this angle line, mark point \(C\) such that \(BC = 6\,\text{cm}\).
- Join \(A\) to \(C\). Now, \(\triangle ABC\) is drawn.
- Draw a helper ray.
- From point \(B\), draw a ray \(BX\) making an acute angle (less than \(90^\circ\)) with side \(BC\).
- Mark equal divisions.
- On \(BX\), use a compass or scale to cut 7 equal parts. Name them \(B_1, B_2, ..., B_7\) starting from \(B\).
- So, \(B = B_0\), then \(B_1, B_2,...,B_7\) are equally spaced points.
- Connect the last point to C.
- Join \(B_7\) to \(C\).
- Use parallel lines to reduce.
- Locate \(B_5\) (because the scale factor is \(\tfrac{5}{7}\)).
- Draw a line from \(B_5\) parallel to \(B_7C\). Let this line meet the extended side \(BC\) at \(C'\).
- Construct the reduced A'.
- Through \(C'\), draw a line parallel to \(AC\).
- This new line meets the line through \(B\) parallel to \(BA\) at point \(A'\).
- Complete the reduced triangle.
- Join \(A'\) and \(C'\). The new triangle \(A'BC'\) is the required reduced triangle.
Justification (Why this works)
From the Basic Proportionality Theorem (Thales’ theorem):
\[
\dfrac{BB_5}{BB_7} = \dfrac{5}{7}
\]
This ratio is carried forward to the sides of the new triangle:
- \(BA' / BA = 5/7\)
- \(BC' / BC = 5/7\)
- \(A'C' / AC = 5/7\)
So, each side of \(\triangle A'BC'\) is \(\tfrac{5}{7}\) of the corresponding side of \(\triangle ABC\). Hence, the triangles are similar with scale factor \(\tfrac{5}{7}\).