29. If from any point in the circumference of the circle described round a triangle perpendiculars be drawn to the sides of the triangle, the three points of intersection are in the same straight line.
Let ABC be a triangle, P any point on the circumference of the circumscribing circle; from P draw PD,
PE,PF perpendiculars to the sides BC, CA,AB respectively: D, B, F shall be in the same straight line.
[We will suppose that P is on the arc cut off by AB, on the opposite side from C, and that B is on CA produced through A; the demonstration will only have to be slightly modified for any other figure.]
A circle will go round PEAF (Note on III. 22); therefore the angle PFE is equal to the angle PAE (III. 21). But the angles PAE and PAC are together equal to two right angles (I. 13); and the angles PAC and PBC are together equal to two right angles (III. 22). Therefore the angle PAB is equal to the angle PBC; therefore the angle PFE is equal to the angle PBC.
Again, a circle will go round PFDB {Note on III. 21); therefore the angles PFD and PBD are together equal to two right angles (III. 22). But the angle PBD has been shewn equal to the angle PFE. Therefore the angles PFD and PFE are together equal to two right angles. Therefore EF and FD are in the same straight line.
