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Question
In a triangle ABC, N is a point on AC such that BN ⊥ AC. If BN2 = AN . NC, prove that ∠B = 90°.
Solution
In ΔABC , BN ⊥ AC.
Also, `BN^2=ANxxNC`
We have to prove that `∠ B= 90^o`.
In triangles ABN and BNC, we have
`AB^2=AN^2+BN^2`
`BC^2=BN^2+CN^2`
Adding above two equations, we get
`AB^2+BC^2=AN^2+CN^2+2BN^2`
Since `BN^2= AN.NC`
So,
`AB^2+BC^2=AN^2+CN^2+2BN^2`
`AB^2+BC^2=(AN+NC)^2`
`AB^2+BC^2=AC^2`
Hence `∠ B = 90^o`
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