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Example 10 - ACB = 90 and CD perpendicular AB. Prove BC2/AC2 - Examples

Example 10 - Chapter 6 Class 10 Triangles - Part 2
Example 10 - Chapter 6 Class 10 Triangles - Part 3

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Example 10 In figure, ∠ ACB = 90° and CD ⊥ AB. Prove that BC2/AC2=BD/AD Given:- ΔCAB, ∠ ACB = 90° And CD ⊥ AB To Prove :- BC2/AC2=BD/AD Proof : From theorem 6.7, If a perpendicular is drawn from the vertex of the right angle to the hypotenuse then triangles on both sides of the Perpendicular are similar to the whole triangle and to each other So, ∆ 𝐴𝐶𝐷 ~ ∆ 𝐴𝐵𝐶 & ∆ 𝐵𝐶𝐷 ~ ∆ 𝐵𝐴𝐶. From (1) ∆ 𝐴𝐶𝐷 ~ ∆ 𝐴𝐵𝐶 If two triangles are similar , then the ratio of their corresponding sides are equal 𝐴𝐶/𝐴𝐵=𝐴𝐷/𝐴𝐶 AC ×𝐴𝐶=𝐴𝐵×𝐴𝐷 AC2 = AB ×𝐴𝐷 Similarly, from (2) ∆ 𝐵𝐶𝐷 ~ ∆ 𝐵𝐴𝐶 If two triangles are similar , then the ratio of their corresponding sides are equal 𝐵𝐶/𝐵𝐴=𝐵𝐷/𝐵𝐶 BC × BC = BA × BD BC2 = BA × BD Divide equation ((4))/((3)) 𝐵𝐶2/𝐴𝐶2=(𝐵𝐴 × 𝐵𝐷)/(𝐴𝐵 × 𝐴𝐷) 𝐵𝐶2/𝐴𝐶2=𝐵𝐷/𝐴𝐷 Hence proved

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