In the given figure, in ∆ABC, point D on side BC is such that, ∠BAC = ∠ADC. Prove that, CA² = CB × CD

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Solution

Given: ∠BAC = ∠ADC
To prove: CA² = CB × CD
Proof: In ∆ABC and ∆DAC
∠BAC = ∠ADC (Given)
∠C = ∠C (Common)
By AA test of similarity
∆ABC ∼ ∆DAC
$∴ \frac{BC}{AC} = \frac{AC}{DC} (Corresponding sides are proportional) \Rightarrow {AC}^{2} = BC \times DC$
Hence proved.

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In the given figure, in ∆ABC, point D on side BC is such that, ∠BAC = ∠ADC. Prove that, CA2 = CB × CD

Given: ∠BAC = ∠ADC To prove: CA2 = CB × CD Proof: In ∆ABC and ∆DAC ∠BAC = ∠ADC (Given) ∠C = ∠C (Common) By AA test of similarity ∆ABC ∼ ∆DAC ∴BCAC=ACDC Corresponding sides are proportional⇒AC2=BC×DC Hence proved.

In the given figure, in ∆ABC, point D on side BC is such that, ∠BAC = ∠ADC. Prove that, CA² = CB × CD

Given: ∠BAC = ∠ADC
To prove: CA² = CB × CD
Proof: In ∆ABC and ∆DAC
∠BAC = ∠ADC (Given)
∠C = ∠C (Common)
By AA test of similarity
∆ABC ∼ ∆DAC
$∴ \frac{BC}{AC} = \frac{AC}{DC} (Corresponding sides are proportional) \Rightarrow {AC}^{2} = BC \times DC$
Hence proved.