PM perpendicularto AO meeting AO in M and produce it to Pl so thatP^M= PM. Then /\ is the image of P. Now, in the triangles MOP and MOPlt PM is equal to P^Mand MO is common, while the angles at M are right angles.Thus the triangles are equal and OP is equal to OPlt ThusP and /*! both lie on a… … way that P2 is on this circle and that the arc _3P2 is equal tothe arc BP^~. Thus the images lie on the circle A PSAS1, AB1being the points in which AO and BO produced cut it—its cir- LIGHT. [CH. Ill cumference takes the place of the straight line in the first case—and their position is found by…
Heat · Light
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