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Question
The chords AB and CD of the circle intersect at point M in the interior of the same circle then prove that CM × BD = BM × AC
Solution
Given: Chords AB and CD intersect at point M.
To prove: CM × BD = BM × AC
Proof: In ∆AMC and ∆DMB,
∠AMC ≅ ∠DMB ......[Vertically opposite angles]
∠ACD ≅ ∠ABD ......[Angles inscribed in the same arc]
∴ ∆AMC ∼ ∆DMB ......[AA test of similarity]
∴ `"CM"/"BM" = "AC"/"BD"` .....[Corresponding sides of similar triangles]
∴ CM × BD = BM × AC
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