Question

\(D\) is a point on the side \(BC\) of \(\triangle ABC\) such that \(\angle ADC = \angle BAC\). Show that:
\(AC^2 = BC \times DC\)

Answer 1

Step 1: Given information:
We are given that \( \angle ADC = \angle BAC \), which implies that triangles \( \triangle ADC \) and \( \triangle ABC \) are similar by the AA (Angle-Angle) similarity criterion. This means that the corresponding angles of the triangles are equal.

Step 2: Applying the similarity ratio:
Since \( \triangle ADC \sim \triangle ABC \), we can apply the proportionality rule that states that the corresponding sides of similar triangles are proportional. This gives us the following proportion:
\[ \frac{AC}{BC} = \frac{DC}{AC} \]

Step 3: Cross-multiply to find the relationship:
To eliminate the fractions, we cross-multiply the equation:
\[ AC \times AC = BC \times DC \] This simplifies to:
\[ AC^2 = BC \times DC \]

Step 4: Conclusion:
The final result is:
\[ AC^2 = BC \times DC \] This is the required result, which relates the sides of the two similar triangles.