NCERT Solutions
Class 10 - Mathematics - Chapter 6: TRIANGLES - Exercise 6.3

Question 13

Given: In \(\triangle ABC\), point D lies on BC such that \(\angle ADC = \angle BAC\).

To prove: \(CA^2 = CB \cdot CD\).

Step 1: Consider triangles \(\triangle ADC\) and \(\triangle BAC\)

We know:

• \(\angle ADC = \angle BAC\) (given)
• \(\angle ACD = \angle BCA\) (common angle at C)

So we have two pairs of equal angles, hence

\[\triangle ADC \sim \triangle BAC \quad (\text{AA similarity}).\]

Step 2: Write the ratio of corresponding sides

From the similarity, match the vertices:

\(D \leftrightarrow A,\ C \leftrightarrow C,\ A \leftrightarrow B.\)

Thus the corresponding sides are:

\[DC \leftrightarrow AC, \quad AC \leftrightarrow BC, \quad AD \leftrightarrow AB.\]

So we get the proportionality:

\[\frac{DC}{AC} = \frac{AC}{BC}.\]

Step 3: Rearrange to get the required relation

Cross-multiplying,

\[DC \cdot BC = AC^2.\]

That is,

\[\boxed{CA^2 = CB \cdot CD}.\]

Hence proved.