D is a point on the side BC of a triangle ABC such that ADC = BAC. Show that CA2 = CB.CD.
Step-by-Step Solution
Key Concept: Use the fact that D lies on BC to obtain a second equal angle (∠ACD = ∠BCA). With two equal angles, triangles ADC and BAC are similar (AA similarity). From the corresponding sides of similar triangles, derive the proportion \(\frac{CA}{CB}=\frac{CD}{CA}\) which gives the required relation \(CA^{2}=CB\cdot CD\).
1. Identify the second equal angle\
Since D lies on BC, the line CD is the same as the line CB. Hence\
\[\angle ACD = \angle BCA\] (both are the angle between AC and the line BC).\
2. State the given equal angle\
By hypothesis,\
\[\angle ADC = \angle BAC\].\
3. Apply AA similarity\
In triangles \(\triangle ADC\) and \(\triangle BAC\) we have\
\[\angle ADC = \angle BAC \quad\text{and}\quad \angle ACD = \angle BCA.\]\
Therefore, \(\triangle ADC \sim \triangle BAC\) (AA similarity).\
4. Write the correspondence of sides\
From the similarity, the sides opposite the equal angles are proportional:
\[
\frac{AC}{BC}=\frac{AD}{BA}=\frac{CD}{AC}.
\]\
(Here, side \(AC\) of \(\triangle ADC\) is opposite \(\angle ADC\) and corresponds to side \(BC\) of \(\triangle BAC\), etc.)\
5. Extract the required proportion\
Using the first and the third ratios we get\
\[
\frac{AC}{BC}=\frac{CD}{AC}\;\Longrightarrow\; AC^{2}=BC\cdot CD.
\]\
This is exactly the statement to be proved.
6. Conclusion\
Hence, when a point D on BC satisfies \(\angle ADC = \angle BAC\), the lengths satisfy \(CA^{2}=CB\cdot CD\).
Correct Answer: CA² = CB·CD