Finding tangents to a circle with a straightedge There is a geometric construction that I heard years ago and I still haven't figured out why it works despite several attempts.
Playing with pen, paper and GeoGebra makes me confident that it does indeed work.
Could someone explain it to me?
Task:
Given a circle and a point outside it, construct the two tangents to the circle through the point, using only a straightedge.
Solution:
Draw any three different lines through the given point $P$ that intersect the circle twice.
Let $A_1,A_2,B_1,B_2,C_1,C_2$ be the six intersection points, with the same letter corresponding to the same line and the index 1 corresponding to the point closer to $P$.
Let $D$ be the point where the lines $A_1B_2$ and $A_2B_1$ intersect, and similarly $E$ for the lines $B_1C_2$ and $B_2C_1$.
Draw a line through $D$ and $E$.
This line meets the circle at two points, $F$ and $G$.
The tangents are $PF$ and $PG$.

It is a consequence of Pascal's theorem in Projective geometry valid for any conic, the circle is a special case. When separation tends to zero, the secant points  $(OA_1, OA_2)$ where $(A_1, A_2) $ are extremities of chord on circle  which multiply to tangent point $F$ as
$ OA_1 OA_2= OB_1 OB_2=...= OD_1 OD_2= PF^2$ as constant product, a property of the circle.
The locus,chord  $ FG $, is a straight line. As a property of circle it is the harmonic mean of segments $ (OA_1, OA_2) $ as,for all chords e.g.,
$$  2/PF = 1/OA_1 + 1/OA_2. $$
Conclusion:
Pascal line is the chord of harmonic means of a pencil/circle intersection

Pascal Line valid on ellipse also, but I do not know if that is the harmonic locus. (posted Sept 2, 2016).

