#### Question

D is a point on the side BC of ∆ABC such that ∠ADC = ∠BAC. Prove that ` \frac{CA}{CD}=\frac{CB}{CA} or, CA^2 = CB × CD.`

#### Solution 1

In ∆ABC and ∆DAC, we have

∠ADC = ∠BAC and ∠C = ∠C

Therefore, by AA-criterion of similarity, we have

∆ABC ~ ∆DAC

`\Rightarrow \frac{AB}{DA}=\frac{BC}{AC}=\frac{AC}{DC}`

`\Rightarrow \frac{CB}{CA}=\frac{CA}{CD}`

#### Solution 2

In ΔADC and ΔBAC,

∠ADC = ∠BAC (Given)

∠ACD = ∠BCA (Common angle)

∴ ΔADC ∼ ΔBAC (By AA similarity criterion)

We know that corresponding sides of similar triangles are in proportion.

`:. (CA)/(CB) = (CD)/(CA)`

=> CA^{2} = CB x CD

Is there an error in this question or solution?

Solution D is a point on the side BC of ∆ABC such that ∠ADC = ∠BAC. Prove that CA/CD=CB/CAor, CA^2 = CB × CD. Concept: Criteria for Similarity of Triangles.