In ▲ACB, BAC=90°. AD perpendicular to BC. Prove that AD²=BD*CD. Easy Tutorial.

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19. Given a ∆ABC in which ang.A=90° & AD per.to BC. Prove that AD^2 = BD.CD 19. Given a ∆ABC in which ang.A=90° & AD per.to BC. Prove that AD^2 = BD.CD
In the given figure, ∠BCA = 90° and AD ⊥ BC, then: In the given figure, ∠BCA = 90° and AD ⊥ BC, then:
In given Fig. ∠BAC=90° and segment AD⊥BC. Prove that AD^2=BD×DC. In given Fig. ∠BAC=90° and segment AD⊥BC. Prove that AD^2=BD×DC.
𝐼𝑛 ∆𝐴𝐵𝐶 𝐵𝐴𝐶=90° 𝐴𝐷 perpendicular to 𝐵𝐶 . 𝑃𝑟𝑜𝑣𝑒 𝑡ℎ𝑎𝑡 2AD2=𝐵𝐷×𝐷𝐶 𝐼𝑛 ∆𝐴𝐵𝐶 𝐵𝐴𝐶=90° 𝐴𝐷 perpendicular to 𝐵𝐶 . 𝑃𝑟𝑜𝑣𝑒 𝑡ℎ𝑎𝑡 2AD2=𝐵𝐷×𝐷𝐶
In △ABC, if AD ⊥ BC and AD²=BD x DC , then prove that ∠BAC=90° #cbse #cbse10th #maths In △ABC, if AD ⊥ BC and AD²=BD x DC , then prove that ∠BAC=90° #cbse #cbse10th #maths
In figure ∠ACB=90 & CD⊥ AB, Prove that CD²=BD×AD In figure ∠ACB=90 & CD⊥ AB, Prove that CD²=BD×AD
In the given figure ∠ BAC = 90° and AD ⊥ BC, then(a) BD·CD = BC² (b) AB·AC = BC²(c) BD·CD = AD² (d) In the given figure ∠ BAC = 90° and AD ⊥ BC, then(a) BD·CD = BC² (b) AB·AC = BC²(c) BD·CD = AD² (d)
In triangle ABC ,if AD perpendicular to BC and AD² = BD × DC then prove that angle BAC = 90⁰ #cbse In triangle ABC ,if AD perpendicular to BC and AD² = BD × DC then prove that angle BAC = 90⁰ #cbse