Consider two lines D1 and D2 and two intersecting lines PAB and PA'B'
We have the following comparison and definition :
|D1 and D2 parallel
(for any secants)
|D1 and D2 are antiparallel
with respect to the secants
|Triangles PAA' and PBB' are similar
||Triangles PAA' and PB'B are similar
|PA/PB = PA'/PB' (Thales)
or PA/PA' = PB/PB'
|PA/PB' = PA'/PB
or PA/PA' = PB'/PB
|Angles PAA' and PBB' are equal
||Angles PAA' and PB'B are equal
We may omit "with respect to" which secant lines when these are not confusing.
Never forget that D1 and D2 are antiparallel only with respect to (wrt) two given secants !
We note D1 // D2 for parallel and D1 \\ D2 for antiparallel (wrt some secant lines).
We can deduce the following theorems :
Two lines antiparallel to a same third line are parallel
D1 \\ D3 and D2 \\ D3 ⇒ D1 // D2
And also, D1 \\ D3 and D1 // D2 ⇒ D1 \\ D3
Obvious from the angle definitions.
We can then speak of "antiparallel directions" (wrt the secants).
Two lines are antiparallel iff quadrilateral AA'BB' is inscribed
Because two opposite angles sum um to 180░.
Note that the distance relations can be written
PA.PB = PA'.PB' = power of P to the circle AA'B'B.
AA' \\ BB' with respect to (AB, A'B') ⇔ AB \\ A'B' with respect to (AA', BB')
Two symmetric lines from the angle bisector are antiparallel
As triangles PIA and PIB' are equal, hence PA = PB'
Similarily PB = PA' hence PA/PB' = PA'/PB ( = 1)
Two lines AA' and BB' are antiparallel iff midpoints of AA' and BB'
That is angles BPN and MPA' are equal.
That is M lies on the symmedian of PBB'.
Considering the medians of two similar triangles PAA' and PB'B,
we deduce that triangles BPN and PA'M are similar, hence angles in P are equal.
Note : lines are parallel iff P,M,N are in line
Tangents to circumcircle
The tangent to circumcircle in a vertex is antiparallel to opposite side
(with respect to the two other sides)
From the angles : angle of tangent with AB is equal to inscribed angle ACB.