In figure`/_A C B=90o` and `C D_|_A B` . Prove that `(B C^2)/(A C^2)=(B D)/(A D)` ....

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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.
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 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 AD is perpendicular to BC then prove that AB²+CD²=BD²+AC² | class 10 maths In the given figure AD is perpendicular to BC then prove that AB²+CD²=BD²+AC² | class 10 maths
In the given Fig., ∠ACB=90° and CD⊥AB. Prove that CB^2/CA^2=BD/AD. In the given Fig., ∠ACB=90° and CD⊥AB. Prove that CB^2/CA^2=BD/AD.
In a Triangle ABC, angle ACB=90 and CD is perpendicular to AB Prove that CB2/CA2=BD/AD | Triangles In a Triangle ABC, angle ACB=90 and CD is perpendicular to AB Prove that CB2/CA2=BD/AD | Triangles
In the given figure, ∠𝑨𝑪𝑩 = 𝟗𝟎° and 𝑪𝑫 ⊥ 𝑨𝑩, Prove that 𝑪𝑫^𝟐 = 𝑩𝑫 × 𝑨𝑫. In the given figure, ∠𝑨𝑪𝑩 = 𝟗𝟎° and 𝑪𝑫 ⊥ 𝑨𝑩, Prove that 𝑪𝑫^𝟐 = 𝑩𝑫 × 𝑨𝑫.
In a triangle ABC, B=90^@ and D is the mid-oint of BC then prove that AC^2=AD^2+3 CD^2 | 10 | X ... In a triangle ABC, B=90^@ and D is the mid-oint of BC then prove that AC^2=AD^2+3 CD^2 | 10 | X ...