In given Fig. ∠BAC=90° and segment AD⊥BC. Prove that AD^2=BD×DC.

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In the given figure, ∠BCA = 90° and AD ⊥ BC, then: In the given figure, ∠BCA = 90° and AD ⊥ BC, then:
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 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 △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 given figure, ∠BAC=900 and segment AD⊥BC. Prove that AD2=BD×DC. In given figure, ∠BAC=900 and segment AD⊥BC. Prove that AD2=BD×DC.
In ABC, if AD ⊥ BC and AD² = BD × DC , then prove that ∠BAC = 90°. #2025boardexamsolutions In ABC, if AD ⊥ BC and AD² = BD × DC , then prove that ∠BAC = 90°. #2025boardexamsolutions
Q33(b) In ABC, if AD BC and AD2 = BD*DC, then prove that Angle BAC = 90°. Q33(b) In ABC, if AD BC and AD2 = BD*DC, then prove that Angle BAC = 90°.
In ▲ACB, BAC=90°. AD perpendicular to BC. Prove that AD²=BD*CD. Easy Tutorial. In ▲ACB, BAC=90°. AD perpendicular to BC. Prove that AD²=BD*CD. Easy Tutorial.