__Synthetic proof of__

Christopher Bradley's Conjecture

Version 2: §8 renewed.

Many thanks to Rafi for initiating a very

animate and interesting discussion thread in the Hyacinthos

newsgroup, starting with the message #8910 "Quadrilateral

problem."; many thanks also to Jean-Pierre Ehrmann, Charles

Worrall, Alexey Zaslavsky, Peter Scales, Marcello Tarquini,

Alexander Bogomolny, Juan Carlos Salazar, Eric Danneels and

Nikolaos Dergiades for taking part and contributing many

interesting details to this thread. By making use of your

contributions, I have succeeded in finding a synthetic proof of

the result which is known as "Christopher Bradley's

Conjecture" in this forum. However, let me apologize for not

having read a great part of the discussion done before I

subscribed to the CTK Exchange, and hence, I may just repeat what

was already said. Let me also excuse for my notations (they are

completely different from those used in this discussion).

**§1. Triangle lemmata**

Consider a triangle ABC and a point P on its

side AC. Let the incircles of triangles PAB and PBC have centers

X and Y, respectively; the incircle of triangle ABC touches AC at

U. Then:

**Theorem 1.** The points P

and U lie on the circle with diameter XY.

**Theorem 2.** The point U

lies on the internal common tangent of the incircles of

triangles PAB and PBC different from the line BP.

For Theorem 1, Juan Carlos Salazar refers to

the American Mathematical Monthly, but he doesn't remember the

exact citation.

*Proofs.* Here is a restated proof of

Theorem 2 by Rafi:

Let U be the point where the internal common

tangent of the incircles centered at X and Y different from BP

meets AC. We will show that the incircle of triangle ABC touches

AC at U.

Let the internal common tangent of the

incircles of triangles PAB and PBC touch these circles at X' and

Y', respectively. Then, since the internal common tangents to two

circles have equal length, we get X'Y' = X"Y", where

X" and Y" are the points where the incircles of

triangles PAB and PBC touch BP.

Now, this yields

PX" - PY" = X"Y" = X'Y'

= UY' - UX'.

If J and K are the points where the incircles

of triangles PAB and PBC touch AC, then the well-known fact that

the two tangent segments from a point to a circle are equal in

length yields that PX" = PJ, PY" = PK, UX' = UJ, UY' =

UK; hence,

PJ - PK = UK - UJ, i. e.

KJ - 2 PK = KJ - 2 UJ,

and PK = UJ. Thus,

AU = AJ + UJ = AJ + PK

= (AB + AP - BP)/2 + (BP + CP - BC)/2

= (AB + AP + CP - BC)/2

= (AB + AC - BC)/2,

and U is the point of tangency of the incircle

of triangle ABC with the side AC. This proves Theorem 2.

Now, the points X and Y lie on the two angle

bisectors of the angle formed by the lines AC and UX'Y' (because

both of these lines touch the two incircles!). Hence, the angle

XUY is 90°, and U lies on the circle with diameter XY. The same

reasoning (using the line BP instead of UX'Y') shows that P lies

on the circle with diameter XY, too. This proves Theorem 1.

**§2. The configuration**

of Christopher Bradley

A *circumscribed quadrilateral* will

mean a quadrilateral with an incircle.

Now let ABCD be a circumscribed quadrilateral

with the incenter O. The diagonals AC and BD meet at P. Let X, Y,

Z, W be the incenters of triangles PAB, PBC, PCD, PDA,

respectively, and (X), (Y), (Z), (W) be the respective incircles.

Then, Christopher Bradley's conjecture states:

**Theorem 3.** The points X,

Y, Z, W lie on one circle.

This fact was proven, together with its

converse, in <1> and <2>. Here, we don't worry about the converse

and seek a synthetic proof.

**§3. Incircles and**

exsimilicenters

Call (O) the incircle of quadrilateral ABCD,

its center being O. Let also (A'), (B'), (C'), (D') be the

incircles of triangles DAB, ABC, BCD, CDA, where A', B', C', D'

are their respective centers.

Since incenters lie on angle bisectors, the

following triples of points are collinear:

(A; O; A'), (B; O; B'), (C; O; C'), (D; O;

D');

(A; B'; X), (B; C'; Y), (C; D'; Z), (D; A'; W);

(A; D'; W), (B; A'; X), (C; B'; Y), (D; C'; Z).

We will use the abbreviation *exsimilicenter*

for the external center of similtude (= external homothetic

center) of two circles.

The *1st Monge theorem* states that the

pairwise exsimilicenters of three circles are collinear.

Let R be the exsimilicenter of the circles (A')

and (C'). Of course, R lies on A'C'. Now, the exsimilicenter of

(O) and (A') is A, and the exsimilicenter of (O) and (C') is C.

Hence, the 1st Monge theorem yields that R lies on the line AC.

The lines AX and CY meet at B'; the lines XA'

and YC' meet at B; the lines AA' and CC' meet at O. Since B, O,

B' are collinear, we can apply the Desargues theorem to triangles

AXA' and CYC', and infer that the lines AC, XY and A'C' concur.

Since AC and A'C' meet at R, we see that R lies on XY.

Now, what have we proven? R lies on the line XY

- joining the centers of the circles (X) and (Y) - and on their

common tangent AC. Hence, R is the exsimilicenter of (X) and (Y).

Similarly, R is the exsimilicenter of (Z) and (W) and lies on the

line ZW. Moreover, there will be a similar point Q on BD with

analogous properties. We have obtained:

**Theorem 4.** The

exsimilicenter R of the circles (A') and (C') is

simultaneously the exsimilicenter of (X) and (Y) and the

exsimilicenter of (Z) and (W). It lies on the lines AC, A'C',

XY and ZW.

The exsimilicenter Q of the circles (B')

and (D') is simultaneously the exsimilicenter of (Y) and (Z)

and the exsimilicenter of (W) and (X). It lies on the lines

BD, B'D', YZ and WX.

Finally, we state a trivial fact which hardly

merits the name "theorem":

**Theorem 5.** The lines XZ

and YW are the angle bisectors of the angles APB = CPD and

DPA = BPC and pass through P.

This is because incenters of triangles lie on

angle bisectors.

**§4. U and T; Proof of**

Theorem 3

We will make use of a well-known result about

circumscribed quadrilaterals:

**Theorem 6.** The incircles

(A') and (C') of triangles DAB and BCD touch BD at one point

T.

The incircles (B') and (D') of triangles

ABC and CDA touch AC at one point U.

Doubters can perform an easy computation of

lengths (making use of the fact that AB + CD = BC + DA in the

circumscribed quadrilateral ABCD).

Of course, the point T lies on A'C'; moreover,

the points T and R divide the segment A'C' harmonically, since T

is the internal center of similtude of (A') and (C') and R is

their external center of similtude.

Now, we can apply Theorem 1 to triangle ABC and

the point P on its side AC. Consequently, the points P and U lie

on the circle with diameter XY. Hence, the points X, Y, P, U are

concylic, and RX * RY = RP * RU. Similarly, RZ * RW = RP * RU,

and thus RX * RY = RZ * RW, and the points X, Y, Z, W are

concyclic. This proves Theorem 3. The last part of this elegant

proof is due to Juan Carlos Salazar.

**§5. The center of the**

circle XYZW

Now regard the center of the circle through X,

Y, Z, W. How could it be else:

**Theorem 7.** The center O'

of the circle through X, Y, Z, W is the point of intersection

of the lines A'C' and B'D'.

This was found by Charles Worrall. Here is,

again, my proof, based on a lemma found by Rafi:

**Theorem 8.** The triangle

PQR is autopolar with respect to the circle through X, Y, Z,

W.

*Proof.* The *diagonal triangle*

of a quadrilateral is the triangle formed by the intersection of

the diagonals and the two intersections of opposite sidelines.

Now, the triangle PQR is the diagonal triangle

of the cyclic quadrilateral XYZW; hence, it is autopolar with

respect to the circumcircle of this cyclic quadrilateral. (In

fact, it is well-known that the diagonal triangle of a cyclic

quadrilateral is autopolar with respect to the circumcircle.)

Theorem 8 is established.

Hence, the line PR - i. e., the line AC - is

the polar of Q with respect to the circle through X, Y, Z, W.

Consequently, the lines AC and O'Q are perpendicular (O' is the

center of the circle through X, Y, Z, W). On the other hand, the

line B'D' is perpendicular to AC, too (since the circles (B') and

(D') touch AC at one point U), and it also passes through Q.

Hence, the line O'Q coincides with the line B'D', i. e. the point

O' lies on B'D'. Similarly, O' lies on A'C'. This proves Theorem

7.

**§6. Another property**

of A'C' and B'D'

There is more to say about the lines A'C' and

B'D':

**Theorem 9.** The line B'D'

is the reflection of the line BD in the angle bisector of the

angle formed by the lines WX and YZ.

The line A'C' is the reflection of the line

AC in the angle bisector of the angle formed by the lines XY

and ZW.

Instead of proving this directly, we will show

a stronger result:

**Theorem 10.** The points P

and O' are isogonal conjugates with respect to any of the

four triangles XQY, YRZ, ZQW, WRX.

*Proof.* Since O' is the circumcenter of

triangle ZWX, we have angle O'WZ = 90° - angle WXZ. On the other

hand, the right-angled triangle WPX gives 90° - angle WXZ = 90°

- angle WXP = angle XWP. Hence, angle O'WZ = angle XWP, and the

line WO' is the reflection of the line WP in the angle bisector

of the angle QWZ. Similarly, the line ZO' is the reflection of

the line ZP in the angle bisector of the angle QZW. In other

words, the reflections of two cevians of P in triangle ZQW in the

corresponding angle bisectors pass through O'. Hence, O' is the

isogonal conjugate of P with respect to triangle ZQW. Similarly

for the triangles XQY, YRZ, WRX.

**§7. Four internal**

common tangents

The geometry related to our configuration

doesn't seem to have an end. If we consider our triangle ABC with

the point P on AC again, and apply Theorem 2, then we see that

the point U lies on the internal common tangent of the incircles

of triangles PAB and PBC different from the line BP. In our

terminology:

**Theorem 11.** The point U

lies on the internal common tangent of the circles (X) and

(Y) different from the line BD.

Similarly we find three other internal

common tangents of (Y) and (Z), of (Z) and (W), and of (W)

and (X).

Let the internal common tangent of the circles

(X) and (Y) different from BD be denoted by b', and the three

other ones by c', d', a' (in the natural order). Then, U lies on

b' and d', and T lies on a' and c'. Now, I conjecture that

**Theorem 12.** The lines a',

b', c', d' are equidistant from O'. In other words, there

exists a circle centered at O' and touching the lines a', b',

c', d'.

Moreover, Peter Scales suspects the following:

**Theorem 13. **The point of

intersection of the lines a' and b' lies on O'X.

Similarly for b' and c', c' and d', d' and

a'.

Proofs of Theorem 12 and 13 would be greatly

appreciated.

**§8. Alexey Zaslavsky's**

tangential quadrangle

In <2>, the editors mention an additional

result proved by Alexey Zaslavsky:

**Theorem 14.** The tangents

to the circle through X, Y, Z, W at X, Y, Z, W form a

quadrilateral with two vertices on AC and two vertices on BD.

Zaslavsky's proof is analytic; here is a

synthetic *proof* based upon an idea of Rafi:

Call (O') the circle through X, Y, Z, W. The

tangents to (O') at W and at X meet at a point A" which is

the pole of the line WX with respect to (O'). Similarly, the

tangents to (O') at X and at Y meet at the pole B" of XY,

the tangents to (O') at Y and at Z meet at the pole C" of

YZ, and the tangents to (O') at Z and at W meet at the pole

D" of ZW.

For establishing Theorem 14, we have to show

that A" and C" lie on AC, and B" and D" lie

on BD.

Since XY passes through R, the pole B" of

XY lies on the polar of R, i. e. on the line PQ (since triangle

PQR is autopolar), i. e. on the line BD. Similarly, D" lies

on BD, and A" and C" on AC. Theorem 14 is proven.

Rafi has lately found out more about these

points A", B", C", D":

**Theorem 14a.** The

quadrilateral A"B"C"D" is inscribed; its

circumcenter lies on the line PO'.

*Proof.* If A_{1}, B_{1},

C_{1}, D_{1} are the midpoints of the segments

WX, XY, YZ, ZW, then, since A", B", C", D"

are the poles of the lines WX, XY, YZ, ZW with respect to (O'),

the points A", B", C", D" are the images of A_{1},

B_{1}, C_{1}, D_{1} in the inversion with

respect to (O').

Now, since A_{1}, B_{1}, C_{1},

D_{1} are the midpoints of the sides of quadrilateral

XYZW, the quadrilateral A_{1}B_{1}C_{1}D_{1}

is a parallelogram with sidelines parallel to the diagonals of

XYZW (Varignon parallelogram), i. e. a rectangle (since the

diagonals XZ and YW of XYZW are perpendicular to each other). The

center of this rectangle A_{1}B_{1}C_{1}D_{1}

is the centroid of the quadrilateral XYZW.

Now, remember that in any inscribed

quadrilateral with perpendicular diagonals, the intersection of

the diagonals is the anticenter (the point of concurrence of the

so-called *maltitudes*, the perpendiculars from the

midpoints of the sides to the opposite sides). Hence, in our

inscribed quadrilateral XYZW with perpendicular diagonals, the

point P is the anticenter. Now, the anticenter of a cyclic

quadrilateral is the reflection of the circumcenter in the

centroid; hence, the centroid of the quadrilateral is the

midpoint between the anticenter and the circumcenter. In our

case, the centroid of the quadrilateral XYZW is the midpoint

between the anticenter P and the circumcenter O'.

In other words, the midpoint O_{1} of

PO' is the center of the rectangle A_{1}B_{1}C_{1}D_{1}.

But any rectangle has a circumcircle centered at the rectangle's

center; hence, the points A_{1}, B_{1}, C_{1},

D_{1} lie on a circle centered at O_{1}.

Now I will use the following fact: Inversions

map circles to circles, but (in general) centers not to centers.

However, the center of the image circle always lies on the line

joining the center of the original circle with the center of

inversion.

Applied to the inversion with respect to the

circle (O'), and remembering that A", B", C",

D" are the inversive images of A_{1}, B_{1},

C_{1}, D_{1}, our result that the points A_{1},

B_{1}, C_{1}, D_{1} lie on a circle

centered at O_{1} implies that the points A",

B", C", D" lie on one circle with a center on the

line O_{1}O', i. e. on the line PO'. This proves Theorem

14a.

**§9. Metric relations**

Last, but not least, there is a series of

metric relations in the configuration. If r, x, y, z, w, a', b',

c', d' are the radii of the circles (O), (X), (Y), (Z), (W),

(A'), (B'), (C'), (D'), respectively, then <3> states that

**Theorem 15.** We have 1/x +

1/z = 1/y + 1/w.

The two proofs given in <3> are not really easy

and little delightful; the following one I have found seems to be

rather nice:

We will establish more than Theorem 15; in

fact, we will show that

**Theorem 16.** We have 1/x -

1/a' = 1/y - 1/c'.

This entails 1/x - 1/y = 1/a' - 1/c', and

similarly, 1/w - 1/z = 1/a' - 1/c'. Therefore, 1/x - 1/y = 1/w -

1/z, proving Theorem 15.

*Proof of Theorem 16.* The points X, Y,

R being collinear, the Menelaos theorem in triangle BC'A' entails

-1 = BX / XA' * A'R / RC' * C'Y / YB (with

signed lengths).

Now, it is easy to prove that BX / XA' = x /

(a' - x) and C'Y / YB = (c' - y) / y; further, A'R / R'C = - a' /

c' (since R is the exsimilicenter of (A') and (C')). Hence,

-1 = x / (a' - x) * (- a' / c') * (c' - y)

/ y,

and

x a' (c' - y) = (a' - x) c' y.

This can be easily seen equivalent to 1/x -

1/a' = 1/y - 1/c', proving our Theorem 15.

**§10. Note about**

excenters

What is true for an incenter also holds for

every excenter - this is the extraversion principle for triangles

(applicable to equations, of course, not to inequalities).

However, whether the incenters of triangles PAB, PBC, PCD, PDA

can be replaced by the excenters corresponding to P in our

configuration *without the need of replacing the circumscribed*

quadrilateral ABCD by a quadrilateral with an excircle is

not clear to me.

Anyway, the analogue of Theorem 3 for excenters

is valid:

**Theorem 17.** The excenters

of triangles PAB, PBC, PCD, PDA corresponding to P lie on one

circle.

This was established by Nikolaos Dergiades.

**References**

<1> Toshio Seimiya, Peter Y. Woo, *Solution*

of Problem 2338, Crux Mathematicorum 4/25 (1999) pp.

243-245.

See also https://www-students.biola.edu/~woopy/math/cr2338.htm

<2> I. Vaynshtejn, *Reshenie zadachi M1524*,

Kvant 3/1996 pp. 25-26.

https://kvant.mccme.ru/1996/03/resheniya_zadachnika_kvanta_ma.htm";

<3> I. Vaynshtejn, N. Vasiljev, V. Senderov, *Reshenie*

zadachi M1495, Kvant 6/1995 pp. 27-28.

https://kvant.mccme.ru/1995/06/resheniya_zadachnika_kvanta_ma.htm

*Darij Grinberg*

*11 Jan 2004*