Question:
In a triangle $A B C, N$ is a point on $A C$ such that $B N \perp A C$. If $B N^{2}=A N$. NC, prove that $\angle B=90^{\circ}$.
Solution:
In $\triangle A B C, B N \perp A C$
Also, $B N^{2}=A N \times N C$
We have to prove that $\angle B=90^{\circ}$.
In triangles ABN and BNC, we have
$A B^{2}=A N^{2}+B N^{2}$
$B C^{2}=B N^{2}+C N^{2}$
Adding above two equations, we get
$A B^{2}+B C^{2}=A N^{2}+C N^{2}+2 B N^{2}$
Since $B N^{2}=A N \cdot N C$
So,
$A B^{2}+B C^{2}=A N^{2}+C N^{2}+2 A N \times N C$
$A B^{2}+B C^{2}=(A N+N C)^{2}$
$A B^{2}+B C^{2}=A C^{2}$
Hence $\angle B=90^{\circ}$
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