Question

D is a point on the side BC of a triangle ABC such that \( \angle ADC = \angle BAC \). Prove that \( CA^2 = CB \times CD \).

Hint:

Whenever two triangles have two equal angles, they are similar — use side ratios to derive relationships between sides.

Answer 1

Step 1: Construction and concept.
In \( \triangle ABC \), draw \( AD \) such that \( \angle ADC = \angle BAC \). We need to prove: \( CA^2 = CB \times CD \).

Step 2: Use the property of similar triangles.
Since \( \angle ADC = \angle BAC \) and \( \angle ACD = \angle ACB \) (common), \[ \triangle CAD \sim \triangle CBA \]
Step 3: Write the ratio of corresponding sides.
\[ \frac{CA}{CB} = \frac{CD}{CA} \]
Step 4: Cross multiply.
\[ CA^2 = CB \times CD \] Step 5: Conclusion.
Hence proved that \( \boxed{CA^2 = CB \times CD} \).