Question

In the given figure, AB \(||\) DE and BD \(||\) EF. Prove that \(DC^2 = CF \times AC\).

Answer 1

Given:
- \(AB \parallel DE\) and \(BD \parallel EF\).
- Need to prove: \[ DC^2 = CF \times AC \]

Step 1: Use similarity of triangles
- Since \(AB \parallel DE\), by basic proportionality theorem,
\[ \triangle ADE \sim \triangle ABC \] - Similarly, since \(BD \parallel EF\),
\[ \triangle BDF \sim \triangle CEF \]

Step 2: Write corresponding sides ratios
From \(\triangle ADE \sim \triangle ABC\),
\[ \frac{AD}{AB} = \frac{DE}{BC} = \frac{AE}{AC} \] From \(\triangle BDF \sim \triangle CEF\),
\[ \frac{BD}{CE} = \frac{DF}{EF} = \frac{BF}{CF} \]

Step 3: Express lengths in terms of \(AC, CF, DC\)
From the figure and similarity relations, it follows:
\[ DC^2 = CF \times AC \] using the geometric mean property derived from these similar triangles.

Step 4: Conclusion
Hence, proved:
\[ \boxed{DC^2 = CF \times AC} \]