Let the semiperimeter of PXC be s. The circle through X and of radius 1 is tangent to AD, so by symmetry the circle centred at A of radius 1 is tangent to PX and is therefore the excircle of PXC opposite P. PB is a tangent from P to this circle. So s = PB = 1. But we also know that the incircle of PXC has radius s - PX = 1 - PX = NP as required.

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