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: Understanding the given information:

We are given that \( \angle ADC = \angle BAC \), which means that triangle \( ADC \) is similar to triangle \( ABC \) by the AA (Angle-Angle) similarity criterion. This tells us that the corresponding angles of the two triangles are equal.

Step 2: Applying the proportionality rule:

Since the triangles are similar, we can apply the proportionality rule for similar triangles. The corresponding sides of similar triangles are proportional. For triangles \( ADC \) and \( ABC \), the proportionality rule is:
\[ \frac{AC}{BC} = \frac{DC}{AC} \]

Step 3: Cross-multiplying:

Now, we can cross-multiply the equation to get rid of the fractions:
\[ AC \times AC = BC \times DC \] \[ AC^2 = BC \times DC \]

Step 4: Conclusion:

Thus, we have derived the required result: \( AC^2 = BC \times DC \).