Solution Tangent Lines to Point Conics Exercise 4.51.
A mathematician is a machine for turning coffee into theorems.
—Paul Erdös (1913–1996)

Exercise 4.51. Complete the case where F = C in the proof of Theorem 4.24.

Included are the proofs for all of the remaining cases. We use the relationships:
I = AB · DE, J = EA · BC, K' = IJ · CD, and F is on AK'.

Case 3. Assume F = C. Note the sets {A, C = F, K'} and {C = F, D, K'} are collinear. Hence {A, C = F, D, K'} is collinear. But this contradicts that A, C, and D are noncollinear.

Case 4.  Assume F = D. Note the sets {A, D = F, K'} and {C, D = F, K'} are collinear. Hence {A, C, D = F, K'} is collinear. But this contradicts that A, C, and D are noncollinear.

Case 5.  Assume F = E. Note the sets {A, E = F, K'}, {A, E = F, J}, {I, J, K'}, and {D, E = F, I} are collinear. Hence {A, D, E = F, I, J, K'} is collinear. But this contradicts that A, D, and E are noncollinear.