In Figure 1, a triangle $ABC$ with $\angle B = 90°$ is shown. Taking $AB$ as diameter, a circle has been drawn intersecting $AC$ at point $P$. Prove that the tangent drawn at point $P$ bisects $BC$.

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Step 1: Let tangent at $P$ meet $BC$ at $R$. $PR = RB$ (tangents from external point $R$ to the circle)... (i)
Step 2: $\angle RPC = \angle RCP$ (proving $\angle RPC = \angle ABP$ as angle in alternate segment, and $\angle ABP = \angle ACP$ as angles in same segment of circle)
Step 3: $\Rightarrow PR = CR$... (ii)
Step 4: From (i) and (ii): $RB = RC$. Hence the tangent at $P$ bisects $BC$.

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