Question

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

Hint:

In geometry proofs involving equal angles, check for AA similarity — it often helps establish proportional sides.

Answer 1

Step 1: Given condition.
In $\triangle ABC$, $D$ lies on $BC$ such that $\angle ADC = \angle BAC$.

Step 2: Draw auxiliary lines.
Join $AD$ and $CA$.

Step 3: Observe similar triangles.
In $\triangle CAD$ and $\triangle CBA$:
\[ \angle ADC = \angle BAC \quad (\text{given}) \]
\[ \angle ACD = \angle ACB \quad (\text{common}) \]
Hence,
\[ \triangle CAD \sim \triangle CBA \quad \text{(by AA similarity)} \]

Step 4: Write the ratio of corresponding sides.
\[ \dfrac{CA}{CB} = \dfrac{CD}{CA} \]

Step 5: Cross-multiply.
\[ CA^2 = CB \times CD \]

Step 6: Conclusion.
Hence proved that $CA^2 = CB \cdot CD$.