In the given figure, ∠𝑨𝑪𝑩 = 𝟗𝟎° and 𝑪𝑫 ⊥ 𝑨𝑩, Prove that 𝑪𝑫^𝟐 = 𝑩𝑫 × 𝑨𝑫.

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In figure ∠ACB=90 & CD⊥ AB, Prove that CD²=BD×AD In figure ∠ACB=90 & CD⊥ AB, Prove that CD²=BD×AD
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In the figure angle ACB = 90⁰ and CD is perpendicular to AB. Prove that BC²/AC²= BD/AD#maths #class In the figure angle ACB = 90⁰ and CD is perpendicular to AB. Prove that BC²/AC²= BD/AD#maths #class
In Tamil, In the figure ∠ACB = 90° and CD⊥AB, prove that CD² = BD × AD | Triangle 6| class 10 ncert In Tamil, In the figure ∠ACB = 90° and CD⊥AB, prove that CD² = BD × AD | Triangle 6| class 10 ncert
In triangle ACB, angle C = 90⁰ CD is perpendicular to AB, then prove that BC²/AC² = BD/AD In triangle ACB, angle C = 90⁰ CD is perpendicular to AB, then prove that BC²/AC² = BD/AD
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In figure`/_A C B=90o` and `C D_|_A B` . Prove that `(B C^2)/(A C^2)=(B D)/(A D)` .... In figure`/_A C B=90o` and `C D_|_A B` . Prove that `(B C^2)/(A C^2)=(B D)/(A D)` ....
In the given figure, ∠ A C B=90^∘ and C D ⊥ A B. Prove that B C^2/A C^2=B D/A D. In the given figure, ∠ A C B=90^∘ and C D ⊥ A B. Prove that B C^2/A C^2=B D/A D.