Finding tangents to a circle with a straightedge

Question

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 $PG$.

Answer 1

The answer is because when you fix one line through $P$ that intersects the circle at $A_1$ and $A_2$ and vary a second line and hence its intersection points $B_1$ and $B_2$, the point $E$ defined by you varies on a line.

This should be fairly easy and boring to check with analytic geometry.

It bugs me too that I cannot see the reason in some ratios or some such though.

Answer 2

EDIT1:

Harmonic mean $HM$ segments are of length $$ (OA_1, OA_2) = OH_A,\, (OB_1, OB_2) = OH_B,\,(OC_1, OC_2)= OH_C,..$$ measured from the pole $O$ always lie on the polar line represented by red circle locus. It contains points $ (H_A, H_B, H_C..) $ and end tangent points $ T_1 T_2 $ where the power of the circle is $ OT_1^2=OT_2^2$.

It is interesting that the neighboring segments can be used as well for this purpose as the points of intersection (black circles) continue to be on the polar, a fact interestingly relevant to draw chord line as tangent points $ T_1 T_2 $ connecting line.

Answer 3

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).