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Problem: In scalene triangle $ABC$, let $K$ be the intersection of the angle bisector of $∠A$ and the perpendicular bisector of $BC$. Prove that the points $A, B, C, K$ are concyclic.

My approach:

enter image description here

My approach is similar to the answer given in this post. I solved it successfully by that using similar method mentioned in the post. But I assumed that $K'$ is the intersection of $(ABC)$ and the side bisector of $BC$ instead of the angle bisector of $\angle A$.

Seeking simplicity: I was wondering that if this problem can be solved directly. Can I use the intersection of the angle bisector and the side bisector(Denoted as $K$) at the beginning of my proof and then somehow show that $K$ lies on $(ABC)$? I want the proof to be direct. I tried hard but I didn't find anything. I don't want to assume angle bisector or the side bisector meet at $K'$ and then show $K=K'$. This feels a bit indirect way to approach. At a first glance, most of will try to construct $K$ and then directly prove it. I tried then failed. Then, the indirect proof seemed easy and I tried it, also got it.

The intended solution according to me: I have collected this problem from the book "Euclidean Geometry in Mathematical OLympiads" by Evan Chen. There are $2$ hints. The $1$st hint says ,"Let $K'$ denote the intersection of the circumcircle and the angle bisector". This seems the author wants to prove $K=K'$ which is an indirect approach. The $2$nd hint says,"Apply lemma $1.18$". That is the incenter/excenter lemma. I think it is for proving $BK'=CK'$ but this can be proved by angle chasing. So, according to me, the intended solution is the indirect approach.

Conclusion: I want a direct solution to this problem(Not proving $K=K'$). Any suggestion is appreciated.

Source: "Euclidean Geometry in Mathematical OLympiads" by Evan Chen(Problem $1.34$)

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3 Answers 3

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Suppose that $\overline{AB} < \overline{AC}.$ Diagram Let $O$ be the circumcentre of $\triangle ABC.$

Let $$\angle OBC = \angle OCB = \alpha$$ $$\angle OAC = \angle OCA = \beta$$ $$\angle OAB = \angle OBA = \gamma$$

Now, $$ \begin{align} \angle AOK &= \angle AOB + \angle BOK \\& = ( 180^\circ- 2 \gamma) + (90^\circ-\alpha)\end{align}$$

Also, $$\begin{align} \angle OAK&=\gamma -\frac{\beta+\gamma}{2} \\&=\frac{\gamma-\beta}{2}\end{align}$$

Using $\,\alpha + \beta + \gamma = 90^\circ,$ it's easy to see that $$\angle OAK = \frac{180^\circ -\angle AOK}{2}$$

Hence, $\triangle AOK$ is isosceles with $OA = OK.$

Since, $OA = OB = OC = OK,$

$A,B,C,K$ are concyclic.

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  • $\begingroup$ You did the indirect approach. I explicitly said that I don't want to show $K=K'$. I want a direct alternate proof. $\endgroup$ Commented May 8, 2025 at 13:55
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    $\begingroup$ I've changed the answer. $\endgroup$ Commented May 8, 2025 at 14:54
  • $\begingroup$ Now, your proof seems to be correct. What was your intuition? How did you come up with the constructions and equations? $\endgroup$ Commented May 8, 2025 at 16:34
  • $\begingroup$ @Math12 Since I knew that $A, B, C, K$ are going to be concyclic and $\triangle AOK$ isosceles, I realized that an all-angle approach would work, especially because of the elegant implication of the equality of two sides from the equality of two angles of an isosceles triangle. $\endgroup$ Commented May 8, 2025 at 16:51
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enter image description here Suppose that $AB \lt AC$.

Extend $AB$ to $D$ such that $AD=AC$.

Draw the red lines as shown.

From $\Delta AKD \cong \Delta AKC$ and $MK$ is the perpendicular bisector of $BC$, we have $$BK=KC=KD$$

We also have

$$\angle KBD=\angle KDB=\angle ACK=C+x $$

Therefore $$\angle DBC=C+2x$$

From exterior angle of triangle,

$$\angle DBC=C+A$$ $$\therefore C+2x=C+A$$ $$x=\frac{A}{2}$$ $$\angle KBC = \angle KAC$$

$\therefore A, B, K, C$ are concyclic.

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  • $\begingroup$ How did you come up with the idea of constructing $D$ such that $AD=AC$? What was your intuition? $\endgroup$ Commented May 8, 2025 at 16:26
  • $\begingroup$ Can you please share your intuition? $\endgroup$ Commented May 15, 2025 at 13:02
  • $\begingroup$ @Math12 Honestly no special intuition. It just occurred to me that I can extend $AB$ to $D$ so as to make two congruent triangles and see if this help. Sorry for the late reply. $\endgroup$ Commented May 15, 2025 at 13:07
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excentral triangle

I have a proof using nine-point circle and another.

Let $I$ be the incenter, $D, E, F$ be the $A$-excenter, $B$-excenter, $C$-excenter of triangle $ABC$.

Since $A, B, C$ are the feets of the altitudes of triangles $DEF$ then the circumcircle of triangle $ABC$ is the nine-point circle of triangle $DEF$.

The nine-point circle passes through the midpoint of $ID$. Moreover, the perpendicular bisector of $BC$ passes through the midpoint of $ID$ ($B, I, C, D$ lies on the circle of diameter $ID$ as $\angle IDB = \angle ICD = \dfrac{\pi}{2}$) and $AI$ passes through the midpoint of $ID$. Hence $K$ is the midpoint of $ID$, which means $A, B, C, K$ are concyclic.

P.S. I cheated. This proof is still indirect but I didn't have to name the new point.

Here is the 2nd proof. Note that $AB \ne AC$ (otherwise, the angle bisector of $\angle BAC$ and the perpendicular bisector of $BC$ are identical instead of intersecting).

enter image description here

Construct $E$ on $AB$ such that $KA = KE$ and $A \ne E$. Therefore $\angle BEK = \angle BAK = \angle KAC$. According to the law of sine $$ \sin\angle EBK = \dfrac{KE}{KB}\sin\angle BEK = \dfrac{KA}{KC}\sin\angle KAC = \sin\angle ACK $$

so $\angle EBK = \angle ACK$ or $\angle EBK + \angle ACK = \pi$.

If the latter is the case then $\angle ACK = \angle ABK$ and $\angle AKB = \angle AKC$ so $\triangle KAB = \triangle KAC$ (angle-side-angle), which implies $AB = AC$, a contradiction.

Hence $\angle EBK = \angle ACK$, which means $\angle ABK + \angle ACK = \pi$. Thus $ABKC$ is cyclic.

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  • $\begingroup$ How did come up with constructing E and using sine law? What was your intuition? $\endgroup$ Commented May 9, 2025 at 7:33
  • $\begingroup$ I wanted to prove that $\angle KBA + \angle KCA = \pi$. So I thought about using congruent triangles. The law of sine was just a technical tool to work around because I couldn't show the triangles $KBE$ and $KCA$ are congruent. $\endgroup$ Commented May 9, 2025 at 13:56
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    $\begingroup$ In the solution, did you mean $\angle BEK=\angle BAK=\angle KAC$? $\endgroup$ Commented May 10, 2025 at 5:37
  • $\begingroup$ Yes. Thanks for pointing that out. I fixed it. $\endgroup$ Commented May 10, 2025 at 6:09

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