Front Page Titles (by Subject) APPENDIX B.: GEOMETRICAL THEOREM. - An Autobiography, vol. 1
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APPENDIX B.: GEOMETRICAL THEOREM. - Herbert Spencer, An Autobiography, vol. 1 
An Autobiography by Herbert Spencer. Illustrated in Two Volumes. Vol. I (New York: D. Appleton and Company 1904).
Part of: An Autobiography
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[From the “Civil Engineer and Architect’s Journal” for July, 1840.]
I believe that the following curious property of a circle has not hitherto been noticed; or if it has, I am not aware of its existence in any of our works on Geometry.
Let ABCDE be a circle, of which ACD is any given segment: Let any number of triangles ABD, ACD, &c., be drawn in this segment, and let circles be inscribed in these triangles: their centres F, G, &c. are in the arc of a circle, whose centre is at E, the middle of the arc of the opposite segment AED.
Join AF, FD, AG, GD; then since F is the centre of the circle, inscribed in the triangle ABD, the lines AF, FD, bisect the angles BAD, BDA. (Euc. B. 4, P. 4). For a like reason AG, GD, bisect the angles CAD, CDA; hence the angles FAD, FDA, together, are equal to half the angles, BAD, BDA together, and the angles GAD, GDA together, to half the angles CAD, CDA together. Now the angles ABD, ACD, are equal (being in the same segment) therefore the angles BAD, BDA together, are equal to the angles CAD, CDA together, and as the halves of equals are equal, the angles FAD, FDA together are equal to the angles GAD, GDA together; that is in the two triangles AFD, AGD, two angles of the one, are together equal to two angles of the other, and therefore the third angle AFD, is equal to the third angle AGD. The same reasoning will prove that all angles similarly circumstanced to AFD, are also equal to AGD: therefore, the points A, F, G, D, are in an arc of a circle.
Join BF, and produce it to cut the opposite circumference in E and join EA, ED; then because the angle ABE, is equal to the angle DBE, the segment AE, is equal to the segment ED, and the chord AE, to the chord ED. Again the angles ABE, EDA, are equal (being in the same segment), and by construction, the angle ADF is equal to the angle FDB, therefore the whole angle EDF, is equal to the two ABF, FDB, that is to the two FBD, FDB, that is to the exterior angle EFD; therefore the angle EFD, is equal to the angle EDF; consequently EF, is equal to ED, that is to EA. The same reasoning would prove EF to be equal to a line drawn from G, to the point E. Wherefore the point E is the centre of a circle, of which F and G, as also the centres of all other circles similarly inscribed, are in the circumference.