This lecture grew out of an attempt to trace the consequences of putting together two examination questions. The subject-matter of both is the reflections X, Y, Z of a point P in the sides of a triangle ABC. In the first question we have to show that the perpendiculars from A on YZ , from B on ZX, from C on XY meet in P′, the centre of the circle XYZ, and that the relation between P and P′ is mutual. Since AY = AP =AZ, the perpendicular from A to YZ bisects it. Hence this perpendicular passes through the centre of the circle XYZ , as do similarly the other two perpendiculars. Let X′, Y′, Z′ be the reflections of P′ in the sides of ABC. Since BC is the common perpendicular bisector of PX and P′X′, PX′ =P′X , and similarly PY′ = P′Y, PZ′ =P′Z. Hence P is the centre of the circle X′Y′Z′, and the two radii are equal.