# Prove the three lines are concurrent.

Let $$O$$ be the circumcenter of $$\triangle ABC$$ with $$\angle A=60^{\circ}$$, $$P$$ be an arbitary point on the circumcircle of $$\triangle BOC$$, and $$D,E,F$$ be the circumcenters of $$\triangle BPC,\triangle CPA, \triangle APB$$ respectively. Prove $$AD,BE,CF$$ are concurrent.

Some intermediate results:

• $$AD$$ bisects $$\angle BAC$$,and $$OD \perp BC$$;
• $$O,P$$ are isogonal conjugate points of $$\triangle DEF$$.

## 2 Answers

Let us denote by $$A',B',C$$ respectively the intersections of $$AD$$, $$BE$$, $$CF$$ with the sides $$BC$$, $$CA$$, $$AB$$.

We have also denoted by $$a_1,a_2;b_1,b_2,c_1,c_2$$ the (lengths of the) angles delimited by lines $$AA'D$$, $$BB'E$$, $$CC'F$$ from $$\hat A, \hat B,\hat C$$. Then we have the relations: \begin{aligned} a_1+a_2 &=\hat A =60^\circ\ ,\\ b_1+c_2 &=180^\circ-\widehat{BPC}=180^\circ-\widehat{BOC}=180^\circ-2\hat A % =180^\circ-120^\circ\\ & =60^\circ\ ,\\ b_2+c_1 &= 180^\circ-a_1-a_2-b_1-c_2 =60^\circ\ . \end{aligned} Then working in the circle $$(F)=(ABP)$$ we have $$\widehat{AFB} =\overset\frown{APB} =\overset\frown{AP} +\overset\frown{PB} =2b_2+2a_1\ .$$ So $$\widehat{FAB}=90^\circ-\frac 12\widehat{AFB}=90^\circ-a_1-b_2$$, giving $$\widehat{FAC}=90^\circ+a_2-b_2$$.

Similarly, $$\widehat{FBC}=(90^\circ-a_1-b_2)+(b_1+b_2)=90^\circ+b_1-a_1$$.

Let us find a formula for the proportion $$C'A:C'B$$ using (only) the above data. \begin{aligned} \frac{C'A}{C'B} &= \frac {\operatorname{Area}AFC} {\operatorname{Area}BFC} = \frac {AF\cdot AC\cdot \sin \widehat{FAC}} {BF\cdot BC\cdot \sin \widehat{FBC}} = \frac {AC\cdot \sin (90^\circ+a_2-b_2)} {BC\cdot \sin (90^\circ+b_1-a_1)} \ . \\[3mm] &\qquad\text{ Similarly:} \\[3mm] \frac{B'A}{B'C} &= \frac {\operatorname{Area}AEB} {\operatorname{Area}CEB} = \frac {AE\cdot AB\cdot \sin \widehat{EAB}} {CE\cdot CB\cdot \sin \widehat{ECB}} =\frac {AB\cdot \sin (90^\circ+a_1-c_1)} {CB\cdot \sin (90^\circ+c_2-a_2)} \ . \end{aligned} As already observed in the OP, the point $$D$$ is on the angle bisector of $$\hat A$$, which gives the third needed proportion $$D'A:D'B$$. Putting all together, and with omission of signs, \begin{aligned} \frac{B'C}{B'A}\cdot \frac{C'A}{C'B}\cdot \frac{A'B}{A'C} &= \frac {BC\cdot \sin (90^\circ+c_2-a_2)} {AB\cdot \sin (90^\circ+a_1-c_1)} \cdot \frac {AC\cdot \sin (90^\circ+a_2-b_2)} {BC\cdot \sin (90^\circ+b_1-a_1)} \cdot \frac {AB} {AC} \\ &= \frac {\sin (90^\circ+c_2-a_2)} {\sin (90^\circ+a_1-c_1)} \cdot \frac {\sin (90^\circ+a_2-b_2)} {\sin (90^\circ+b_1-a_1)} \\ &=1\ , \end{aligned} because we can express all angles in terms of $$a_1,b_1,c_1$$, and then \begin{aligned} c_2-a_2=(60^\circ-b_1)-(60^\circ-a_1)=a_1-b_1\ ,\\ a_2-b_2=(60^\circ-a_1)-(60^\circ-c_1)=c_1-a_1\ , \end{aligned} (and since the two values $$\sin(90\pm ?)$$ coincide,) we obtain the cancellations of sine factors.

Now use the reciprocal of the theorem of Ceva to obtain $$AA'$$, $$BB'$$, $$CC'$$ concurrent.

$$\square$$

Another way to prove this is to use projective geometry maps.

Observe that $$E$$ and $$F$$ are on fixed lines (they lie on perpendicular segment bisector for $$AC$$ and $$AB$$ respectively, name them $$e$$ and $$f$$). So when moving $$P$$ on circle $$BOC$$ we see that angle $$\angle FDE$$ is constant ($$=60^{\circ}$$), so the transformation $$DE\mapsto DF$$ is a projective map (induced by rotation around $$D$$ for $$60^{\circ}$$) of pencil of lines through $$D$$ to it self. This one induce new projective map $$E\mapsto F$$ from $$e$$ to $$f$$. But now we have new projective map from a pencil of lines through $$B$$ to a pencil of lines through $$C$$: $$\pi: BE\longmapsto CF$$ and let $$X$$ be their intersection point. Since $$BC$$ goes to it self so $$\pi$$ is perspective so $$X$$ describes some line. So we need only two position of $$P$$ to show that $$X$$ is on $$AD$$ and then you arte done.

Now this you could easly check if you put say $$P=I$$ and $$P=O$$ and you are done.

• I think, for the problem, what we need to do is only show that $BA,BE,BC,BD$ and $CA,CF,CB,CD$ are perspective. Commented Feb 28, 2020 at 14:26
• And, since $BC$ and $CB$ are corresponding to itself, so we only need to show the two pencils above are projective. Commented Feb 28, 2020 at 14:32
• Yes that is correct, I see now, if you put $P=I$ is also obvious. Commented Feb 28, 2020 at 14:44
• But I fail to comprehend well what you said, say, "But for this we need only three differnt positions for $E$ ". Commented Feb 28, 2020 at 14:58
• Yes, you are right, if $BC$ goes to it self then $X$ describes some line, so we need only two position of $P$ to show that $X$ is on $AD$ and then you arte done. Commented Feb 28, 2020 at 15:00