- In Hyacinthos message #9979, Antreas considered the

configuration arising from a triangle ABC, an arbitrary

point P in its plane, and the circumcevian triangle

A'B'C' of the point P. The reflections of the points A',

B', C' in the lines BC, CA, AB were denoted by A*, B*,

C*.

In Hyacinthos message #9983, I wrote about this

configuration:

>> The lines AA*, BB*, CC* concur if and only if

There is actually much more to say about this

>> the point P lies on the line at infinity or on

>> the Jerabek hyperbola. Notably, if P is an

>> infinite point, the point of concurrence of

>> the lines AA*, BB*, CC* seems to be the

>> isogonal conjugate of the point P with respect

>> to the orthic triangle of triangle ABC, and

>> hence should lie on the nine-point circle of

>> triangle ABC.

concurrence. Here is a listing of properties:

(1) Let g be an arbitrary line in the plane of triangle

ABC, and let Ha, Hb, Hc be the feet of the altitudes of

triangle ABC. Let ga, gb, gc be the parallels to the

line g through the points Ha, Hb, Hc, and let a', b', c'

be the reflections of the lines ga, gb, gc in the lines

BC, CA, AB, respectively. Then, the lines a', b', c'

concur at one point V on the nine-point circle of

triangle ABC.

(This is a result by Floor van Lamoen.)

(2) If P is the infinite point of the line g, then the

point V is the isogonal conjugate of P with respect to

triangle HaHbHc.

(And now our configuration enters, with slightly

modified definition:)

(3) If the parallels to the line g through the points A,

B, C intersect the circumcircle of triangle ABC at the

points A', B', C' (apart from the points A, B, C), and

if A*, B*, C* are the reflections of the points A', B',

C' in the lines BC, CA, AB, then the points A*, B*, C*

are the reflections of the points A, B, C in the point

V.

(And this makes clear that the lines AA*, BB*, CC*

concur at V.)

(Now a few off-topic properties, meaning that these

properties do not really involve A*, B*, C*:)

(4) If h is the perpendicular to the line g from the

circumcenter O of triangle ABC, then the triangle A'B'C'

is the reflection of the triangle ABC in the line h.

(5) The lines A'A*, B'B*, C'C*, i. e. the perpendiculars

to the lines BC, CA, AB through the points A', B', C',

concur at a point Q on the circumcircle of triangle ABC.

(6) The Simson line of this point Q with respect to

triangle ABC is parallel to g.

PROOFS.

In the following, we work with directed angles modulo

180; all angles are measured in degrees.

There are two proofs for assertion (1). The first proof

is an angle chase:

Let the line a' meet the nine-point circle of triangle

ABC at a point V (apart from the point Ha); then, if we

succeed to prove that this point V also lies on the

lines b' and c', then the proof of (1) is complete.

Hence, it remains to prove that this point V also lies

on the lines b' and c'.

Since the point V lies on the nine-point circle, we have

< HaVHb = < HaHcHb. In other words,

< (a'; HbV) = < HaHcHb. It is well-known that

< HaHcHb = 2 < BCA, and thus < (a'; HbV) = 2 < BCA.

On the other hand, since the line a' is the reflection

of the line ga in BC, we have < (a'; BC) = < (BC; ga).

Since ga || g, this becomes < (a'; BC) = < (BC; g).

Similarly, < (b'; CA) = < (CA; g), and thus

< (a'; b') = < (a'; BC) + < (BC; CA) - < (b'; CA)

= < (BC; g) + < (BC; CA) - < (CA; g)

= (< (BC; g) - < (CA; g)) + < (BC; CA)

= < (BC; CA) + < (BC; CA)

= 2 < (BC; CA) = 2 < BCA = < (a'; HbV).

Thus, b' || HbV, and since the lines b' and HbV have the

point Hb in common, they must coincide, i. e. the point

V lies on the line b'. Similarly, V also lies on c'.

This completes the first proof of (1).

The second proof of (1) is merely philosophical:

Consider the infinite point P common to the parallel

lines g, ga, gb, gc. The lines ga, gb, gc are the

cevians of this point P with respect to the triangle

HaHbHc. Now, it is well-known that the lines BC, CA, AB

are the angle bisectors (internal or external, this

doesn't matter) of this triangle HaHbHc. Hence, the

lines a', b', c' are the reflections of the cevians of

the point P with respect to the triangle HaHbHc in the

respective angle bisectors; thus, they concur at the

isogonal conjugate of P with respect to triangle HaHbHc.

And since the point P is an infinite point, its isogonal

conjugate lies on the circumcircle of triangle HaHbHc,

i. e. on the nine-point circle of triangle ABC. This

proves (1) again.

The proof of (2) is contained in the second proof of (1)

given above.

The proof of (3) is a kind of angle chase again: We will

show that the point B* is the reflection of B in V. (The

similar assertions about C* and A* will then follow by

analogy.)

In order to prove that the point B* is the reflection of

B in V, we follow the "working backwards" strategy: We

*define* the point B* as the reflection of B in V, then

we *define* the point B' as the reflection of B* in CA,

and then we try to show that the point B' is the point

of intersection of the parallel to g through B with the

circumcircle of triangle ABC.

Consider the midpoints Ma, Mb, Mc of the sides BC, CA,

AB of triangle ABC. The points Ha, Hb, Hc, Ma, Mb, Mc

and V all lie on the nine-point circle of triangle ABC.

Thus, < MaVMc = < MaMbMc; but clearly < MaMbMc = < ABC,

and thus < MaVMc = < ABC.

Since the point B* is the reflection of B in V, the

point V is the midpoint of BB*. On the other hand, the

point Ma is the midpoint of BC. Hence, VMa || B*C.

Similarly, VMc || B*A. Thus,

< (B*C; B*A) = < (VMa; VMc), or, in other words,

< CB*A = < MaVMc. Together with < MaVMc = < ABC, this

becomes < CB*A = < ABC.

Since the point B' is the reflection of B* in CA, we

have < CB'A = - < CB*A = - < ABC = < CBA, and thus, the

point B' lies on the circumcircle of triangle ABC. So

the only thing left to show is that the point B' lies on

the parallel to g through B; once this will be shown,

the proof of (3) will be complete.

Let S be the reflection of B in Hb. Then, Hb is the

midpoint of BS. On the other hand, V is the midpoint of

BB*. Thus, HbV || SB*; in other words, b' || SB*.

Now, since S is the reflection of B in Hb, we can also

say that S is the reflection of B in CA. On the other

hand, B* is the reflection of B' in CA. Thus, the lines

BB' and SB* are symmetric to each other with respect to

the line CA. Also, per definitionem, the lines gb and b'

are symmetric to each other with respect to the line CA.

Since b' || SB*, it follows that gb || BB'. Together

with gb || g, this implies g || BB', and thus, the point

B' lies on the parallel to g through B. This completes

the proof of (3).

(The last part of the proof of (3) could be also done

without introducing S, but doing an angle chase

instead.)

The assertion (4) is trivial: Since AA' is a chord in

the circumcircle of triangle ABC, whose center is O, the

perpendicular bisector of the segment AA' passes through

O. This perpendicular bisector is also perpendicular to

g (since AA' || g); thus, this perpendicular bisector is

the line h, and the points A and A' are symmetric to

each other with respect to h. Similarly, the points B

and B' are symmetric to each other with respect to h,

and the same holds for C and C'. This proves (4).

As for (5), let's call Q the point where the

perpendicular to BC through A' meets the circumcircle of

triangle ABC (apart from A', of course). We will show

that this point Q lies on the perpendiculars to CA and

AB through B' and C', too.

Since Q lies on the circumcircle, we have

< A'QB' = < A'C'B'. Because of (4), the triangles ABC

and A'B'C' are indirectly congruent, so

< A'C'B' = - < ACB. Consequently,

< A'QB' = - < ACB = < BCA. In other words,

< (A'Q; B'Q) = < (BC; CA). Since the point Q lies on the

perpendicular to BC through A', we have

< (BC; A'Q) = 90. Thus,

< (CA; B'Q) = < (CA; BC) + < (BC; A'Q) + < (A'Q; B'Q)

= < (CA; BC) + 90 + < (BC; CA) = 90,

and thus, the point Q also lies on the perpendicular to

CA through B'. Similarly, Q lies on the perpendicular to

AB through C'. This completes the proof of (5).

Last but not least, a proof of (6): The Simson line of

the point Q with respect to triangle ABC passes through

the orthogonal projections X, Y, Z of Q on the sides BC,

CA, AB. We have to prove that YZ || g.

Well, since < AYQ = 90 and < AZQ = 90, the points Y and

Z lie on the circle with diameter AQ, so that

< YZA = < YQA, and consequently,

< (YZ; AB) = < YZA = < YQA = < (QY; AQ)

= < (QY; CA) + < (CA; AQ)

= 90 + < (CA; AQ)

(since QY is perpendicular to CA)

= 90 + < CAQ

= 90 + < CC'Q (cyclic)

= 90 + < (CC'; C'Q)

= 90 + < (g; C'Q) (since CC' || g)

= 90 + < (g; AB) + < (AB; C'Q)

= 90 + < (g; AB) + 90

(since C'Q is perpendicular to AB)

= < (g; AB),

so that YZ || g, and we are done.

Darij Grinberg - Dear Jean-Pierre,

In Hyacinthos message #10581, you wrote:

>> Note that the result remains true if, instead

Indeed, that's clear (it's all about the angles).

>> of the midpoints of the sides, you take for

>> MaMbMc a triangle inscribed in ABC and

>> directly similar with ABC : the lines a',b',c'

>> concur at a point of the circle MaMbMc.

[...]

>> Unfortunately, if instead of three parallel

Yes, of course, it doesn't remain true. Maybe

>> lines ga,gb,gc, we consider three concurrent

>> lines ga,gb,gc, it dosn't seem that the

>> result remains true.

this could give a good locus...

Sincerely,

Darij Grinberg