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: Triangle Similarity
- Because \(AB \parallel DE\), by the basic proportionality theorem, \[ \triangle ADE \sim \triangle ABC \] - Similarly, since \(BD \parallel EF\), \[ \triangle BDF \sim \triangle CEF \]

Step 2: Side 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: Length Relationships
From the figure and similarity, we obtain: \[ DC^2 = CF \times AC \] using the geometric mean property.

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