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1.
Drop perpendicular AD on BC (from A on BC). Now, BD = BC/2. Finally apply similarity by AA in triangles ABD and CBP.
2.
2AP.BP = 2(AB - BP).BP = 2AB.BP - 2BP^2 = BC^2 - 2BP^2. (uses 1)
Using Pythagoras', in CBP, we get, BC^2 = BP^2 + CP^2.
So, 2AP.BP = BC^2 - 2BP^2 = BP^2 + CP^2 - 2BP^2 = CP^2 - BP^2.
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