Let $ABC$ be a triangle inscribed in circle $\omega$, and let the medians from $B$ and $C$ intersect $\omega$ at $D$ and $E$ respectively. Let $O_1$ be the center of the circle through $D$ tangent to $AC$ at $C$, and let $O_2$ be the center of the circle through $E$ tangent to $AB$ at $B$. Prove that $O_1$, $O_2$, and the nine-point center of $ABC$ are collinear.

My Progress:

Here's the diagram enter image description here

Define : $F,X,Y$ as midpoints of $BC,CA,AB$ .

$N_9$ as the nine point center

$O$ as the circumcentre

$H'$ as the orthocentre

$BX \cap (O_1) = L$

$CY \cap (O_2)= I $

Claim: $ABCL$ is a parallelogram

Proof: Since $AX=CX$ by midpoint condition and $BX=XL$ by POP ( Taking power of $X$ wrt both circles $XD.XL=CX^2= XD.XB$ )

Similarly $ABCI$ is a parallelogram

Claim: $ALI$ are collinear

Proof: That follows from BC parallel condition

We also know that $H, M,N_9,O$ are collinear , $M$ is the centroid

Now what we noticed was that $OO_2H'O_1$ are parallelogram with $N_9$ as the intersection of the diagonals .

What I think is that showing $OO_2H'O_1$ a parallelogram is enough , since we know that $N_9$ is the midpoint of OH'

There's also nice dilations happening , like $N_9$ dilating $O$ to $H'$ and $O_2$ to $O_1$ with scale $-1$ ( observation )

dilation centred at $X$ and $Y$ with scale factor -1 too.

Moreover, we also know that $OO_2 \perp BE$ , so it's enough to show that $O_1H' \perp BE$ .

Also I want to find a pure synthetic method ( not using tring, cord , Bary, vector , etc ) but can include inversion or projective .

Thanks in advance!

Here's the diagram link: https://www.geogebra.org/geometry/g3mbkhkp

EDIT: Since @Anand told me to define $IB\cap LC$ , I defined $IB\cap LC=J$

enter image description here

Since $IL || BC$ , $A$ is midpoint of $IA$ and $F$ is midpoint of $BC$ , we get that $JFA$ is collinear and $B,F,C$ are midpoints of $IJ,AJ,LC$

So $AJ,BL,CI$ concur at $K$ , $K$ is the centroid and we get that $K$ is the centre of dilation with scale factor $-2$ .

also $K$ dilates $O$ to $H'$ too .

enter image description here

That's how much I could proceed till now :(

  • $\begingroup$ Is $OO_1H'O_2$ always a parallelogram? There is a configuration in your diagram when it is a concave quadrilateral. $\endgroup$ Sep 7, 2020 at 20:58
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    $\begingroup$ Woah! Really nice observations!! Here's a hint: Aren't your observations begging you to define $IB\cap LC$? $\endgroup$
    – Anand
    Sep 8, 2020 at 5:41
  • $\begingroup$ @Anand actually no , Should I define IB\cap LC ..? $\endgroup$ Sep 8, 2020 at 8:13
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    $\begingroup$ @Anand oh, H' is the centre of that circle , so everything is dilated with centre as G ..Nice! but how is this going to help us ? $\endgroup$ Sep 8, 2020 at 8:52
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    $\begingroup$ Hello Sunaina! It's been over two weeks since i haven't seen one of your informative posts, hope you're doing good. $\endgroup$
    – Arjun
    Sep 23, 2020 at 7:55

1 Answer 1


enter image description here

Here is my way to prove it. My apology if there is some point notation that is different that in your original problem as I only denote the point that is stated in your problem, not your figure. Moreover, there are idea of the problem that is obvious by angles transformation, hence I won't go detail into it.

Now let $G$ is the centroid of $\triangle ABC$, $F$ is midpoint of $BC$;$K, L$ is the intersection point of $BX$ with $O_2$ and $CY$ with $O_1$ respectively. Then denote $J, I$ are intersections point of $BX, CY$ with 9-point circle, respectively. And $T,Z$ are intersection points of $CY$ with $O_2$, $BX$ with $O_1$. Hence we will have: $EYJB$ are cyclic ($\angle BEC = \angle XFY$) which infer that $BE\parallel XI$. Similarly, we obtain $DXIC, DZLC$ are cyclic and $CD \parallel YJ$. Now since $BE\parallel XI$ and $DXIC$ are cyclic, we obtain that $BE\parallel ZL$. By the same way, $CD \parallel TK$.

Up to now, notice that $TZLK$ are cyclic (angle transformation), it follows that if $O_3$ is the center of $(TZLK)$ then $O_3O_2 \perp TK$ or $O_3O_2 \perp CD$. More insightful, $(TZLK)$ is nothing but a homothety of $(XYJI)$ with ratio $\displaystyle \frac{1}{3}$ with homothetic center is $G$. Is this ratio remind you about something? Yes, it is the ratio of $\frac{GN}{GH}=\frac{1}{3}$ where $H$ is the orthocenter of $\triangle ABC$ or we can say $H$ is a homothety of $N$ with the homothetic center is $G$. Therefore $H$ is the center of $(TZLK)$ and $H\equiv O_3$. Finally we have $HO_1 \perp BE$ and $HO_2 \perp CD$ as you desire.

  • $\begingroup$ What are angle transformations? $\endgroup$
    – Anand
    Sep 8, 2020 at 5:58
  • $\begingroup$ Oh it's just dealing with equal angle. $\angle A = \angle B$ and $\angle B = \angle C$ so $\angle A = \angle C$. It's nothing but my english. $\endgroup$ Sep 8, 2020 at 6:00
  • $\begingroup$ So what's the transformation here? Also, what do you mean by "there are idea of the problem that is obvious by angles transformation,"? Sorry for bothering you. $\endgroup$
    – Anand
    Sep 8, 2020 at 6:04
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    $\begingroup$ Oh, got it! Thanks :) $\endgroup$
    – Anand
    Sep 8, 2020 at 6:11
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    $\begingroup$ @NguyễnQuân is it okay, if I read your solution a bit later, I still want to try it on my own .. Thanks! $\endgroup$ Sep 8, 2020 at 8:21

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