**Hubert Shutrick** **January 18, 2012**

Griffith's theorem is proved algebraically using complex numbers elsewhere on ctk and illustrated by an applet. For completeness, the following is a simple geometrical proof that is an adaptation of a proof available on the web. The steps in the proof are illustrated by separate diagrams to make it more visual and less verbal.

**Griffith's Theorem**. Consider the points A',B',C' whose vertices are the midpoints of the sides of the triangle ABC and let round brackets denote the circle through the points they enclose so that (A',B',C') is the nine-point circle of ABC, Corresponding to each line through the centre O of (A,B,C), there is a point G on (A',B',C') such that the pedal circle with respect to ABC of any point P on the line passes through G.

**Proof**. The point where the line through B' orthogonal to AC intersects the circle (A,B',C') is, in fact, O because the right angle gives that AO is the diameter of the circle and \angle AC'O is then also 90°. Consider any line OF through O determined by the other intersection point F between it and (A,B',C') and let G be the image of F under the involution that reflects points in the line B'C'. It will turn out to be the required point of the theorem.

The involution takes A to the foot point A_{h} of the altitude from A to BC so the image of (A,B',C') is (A_{h},B',C'), which is the nine point circle (A',B',C') and G is on it.

For any point P on the line OF, and let A_{p},B_{p},C_{p} be its pedal points and let L be the image of A_{p} under the involution. The quadrilateral AC_{p}FL is cyclic because all the points lie on the circle with diameter AP due to the right angles. Hence,

- \angle C_{p}FL = \angle AC'B'

both being equal to the external angle of the quadrilateral at A.

Let X be the intersection of B_{p}C_{p} and B'C' . The quadrilateral C_{p}C'FX is cyclic, since the angles XC_{p}F and XC'F are both equal to \angle FAC, and it follows that

- \angle C_{p}FX = \angle AC'B'

and so the point X lies on FL as the diagram suggests.

Finally, LX\cdot FX = C_{p}X\cdot B_{p}X since the quadrilateral is cyclic, but LX\cdot FX = A_{p}X\cdot GX from the involution so A_{p}X\cdot GX = C_{p}X\cdot B_{p}X and G lies on (A_{p},B_{p},C_{p}) as required.

**Fontene's theorem** states that (A_{p},B_{p},C_{p}) passes through the point G where A_{p}X intersects (A',B',C'), the same applying to B_{p}Y and C_{p}Z defined similarly.