In the given figure, ∠ B A C=90^∘ and A D ⊥ B C. Then, (a) B C · C D=B C^2 (b) A B · A C=B C^2 (c...

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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 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 the given figure, `angle BAC=90^(@) and AD bot BC`.then, In the given figure, `angle BAC=90^(@) and AD bot 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 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 angle B= 90. AB=BC=2cm. Find AC in the given figure angle B= 90. AB=BC=2cm. Find AC
In the given figure, DEFG is a square and ∠BAC=90°.Show that FG^2=BG×FC. SIMILAR TRIANGLES CLASS-X In the given figure, DEFG is a square and ∠BAC=90°.Show that FG^2=BG×FC. SIMILAR TRIANGLES CLASS-X
How to make angle bisector How to make angle bisector