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".`

D is a point on the side BC of triangle ABC such that angle ADC is equal to angle BAC. Prove that: CA² = CB × CD.

Answer 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"}`

Answer 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")`

⇒ CA² = CB × CD