In the given figure, in ∆ABC, point D on side BC is such that, ∠BAC = ∠ADC. Prove that, CA2 = CB × CD
Practice Set 1.3 | Q 9 | Page 22
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
\[\therefore \frac{BC}{AC} = \frac{AC}{DC} \left( \text{ Corresponding sides are proportional } \right)\]
\[ \Rightarrow {AC}^2 = BC \times DC\]
Hence proved.
Answer:-
Given: ∠BAC = ∠ADC To prove: CA^2 = CB × CD Proof: In ∆ABC and ∆DAC, we have: ∠BAC = ∠ADC … (Given) ∠C = ∠C … (Common angle) By AA test of similarity, we have: ∆ABC ~ ∆DAC So, we can write the following proportion: BC/AC = AC/DC … (Corresponding sides are proportional) This can be simplified to: AC^2 = BC × DC Hence, we have proved that CA^2 = CB × CD.
Chapter 1. Similarity- Practice Set 1.3 – Page 22
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