Question

D is a point on the side BC of $△ A BC$ such that $∠ C A B = ∠ C D A$ . Show that $C A^{2} = CB \times C D$ .

Approach

In triangle ABC, D is on BC such that $∠ C A B = ∠ C D A$ . Since these angles subtend the same arc or are corresponding angles in similar triangles, $△ A D C \sim △ B A C$ . From similarity, $\frac{D C}{A C} = \frac{A C}{BC}$ , hence $C A^{2} = CB \times C D$ .

Step-by-step working

1
Given $∠ C A B = ∠ C D A$ , and $∠ A C D$ is common, so $△ A D C \sim △ B A C$ by AA similarity.
1 mark
2
From similarity: $\frac{D C}{A C} = \frac{A C}{BC}$
0.5 marks
3
Cross-multiply: $A C^{2} = D C \times BC$ or $C A^{2} = CB \times C D$
0.5 marks

Concepts used

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Approach

Step-by-step working

Concepts used

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