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1 III, 5| HP too will be to KP as HI is to IP. But this is also 2 III, 5| MK, for the ratio both of HI to IP and of MH to MK is 3 III, 5| which HM and MI make with HI will always be the same. 4 III, 5| a number of triangles on HI and KI equal to the triangles 5 III, 5| perpendiculars will fall on HI at the same point and will 6 III, 5| horizon. Let the axis now be HI. The proof will be the same 7 III, 5| centre of that circle (namely HI) which now determines the