# Prove that the intersection point of lines $AK$ and $CL$ lies on the line $BO$

$AA', BB'$ and $CC'$ heights of an acute triangle $ABC$. The circle with center $B$ and radius $BB'$ intersects the line $A'C'$ in the points $K$ and $L$. Prove that the intersection point of lines $AK$ and $CL$ lies on the line $BO$, where $O -$ center of the circle circumscribed about the triangle $ABC$.

My wotk so far:

$H -$ orthocenter of triangle $ABC$ is the center of the inscribed circle of triangle $A'B'C'$ • Notice that the diagram contains many symmetric kites which are also cyclic. – Element118 Apr 7 '16 at 15:17
• Please explain why $\angle K$ and $\angle L$ are marked as equal. – Mick Apr 7 '16 at 18:10
• @Mick Explained in my proof. – Oldboy Jun 26 '18 at 4:36
• @Element118 Some kites are obviously cyclic, some are not. Please check my proof. – Oldboy Jun 26 '18 at 4:37

## 1 Answer

Too bad this problem did not get more attention. A lot of things going on here... First, let's prove that $\triangle BB'C\cong \triangle BLC$

$BC'HA'$ is a cyclic quadrilateral so $\angle BA'C'=\angle BHC'=\angle CA'L=\alpha$.

$A'HB'C$ is also a cyclic quadrilateral so $\angle B'A'C=\angle B'HC=\alpha$.

So $\angle B'A'C=\angle LA'C$ which also means that $\angle BA'B'=\angle BA'L$.

Consider triangles $\triangle BA'B'$ and $\triangle BA'L$. They have one common side ($BA'$), a pair of equal sides ($BB'=BL$) and a pair of equal angles ($\angle BA'B'=\angle BA'L$).

Consequentially:

$$\triangle BA'B'\cong \triangle BA'L$$

...which also means that $B'A'=A'L$. Now it's trivial to prove that:

$$\triangle B'A'C\cong \triangle LA'C$$

$$\triangle BB'C\cong \triangle BLC$$

$$\angle BLP=90^\circ$$

Also note that in $\triangle A'CL$:

$$\angle A'CL=\angle A'CB'=\gamma, \space\angle CA'L=\alpha\implies A'LC=\beta$$

You can show in exactly the same way that:

$$\angle BKP=90^\circ, \space \angle C'KA=\beta$$

So triangles $\triangle BKL$ and $\triangle KLP$ are isosceles and quadrilateral $BKPL$ is symmetric with respect to $BP$.

Now let us prove that $\angle ABP=\angle B'BC=90^\circ-\gamma$

In triangle $\triangle ABP:$

$$\angle BAP=180^\circ-\angle BAK=180^\circ-\angle BAC=180^\circ-\alpha$$

$$\angle BPA=90^\circ-\beta$$

$$\angle ABP=180^\circ-\angle BAP - \angle BPA=180^\circ-(180^\circ-\alpha)-(90^\circ-\beta)=90^\circ-\gamma$$

So we proved that:

$$\angle ABP=\angle B'BC$$

It's a well known (and easily provable) fact that bisector of $\angle B$ splits not only the angle itslef but also the angle between the height $BB'$ and the line that contains point $B$ and the center of circumscribed circle $O$ (for more details you can check this page or Wikipedia; if you still want the proof of this fact, please let me know). In other words, lines $BB'$ and $BO$ are symmetric with respect to bisector of $\angle B$.

In this particular case, $\angle ABP=\angle B'BC$ which means that lines $BB'$ and $BP$ are actually symmetric with respect to the bisector of $\angle B$ (not shown in the picture). This means that the line $BP$ must contain the center of circumscribed circle $O$. Done!

EDIT: Mick mentioned that I was using ASS to prove that triangles $\triangle BA'B'$ and $\triangle BA'L$ were congruent. I think that I had the right to do because in acute triangles the angle at $A'$ is always obtuse. Quote from Wikipedia: If two triangles satisfy the SSA condition and the length of the side opposite the angle is greater than or equal to the length of the adjacent side (SSA, or long side-short side-angle), then the two triangles are congruent. But, OK, I admit that ASS is ugly so let's prove the same without it. Maybe not the easiest proof but it still works: So we want to prove (without ASS) that $A'B'=A'L$ knowing that:

$$BB'=BL, \space \angle BA'B'=\angle BA'L>90^\circ$$

Draw circumscribed circles for triangles $\triangle BA'B'$ and $\triangle BA'L$. Central angles $\angle BO_1L$ and $\angle BO_2B'$ are the same because they correspond to the same inscribed angle ($\angle BA'B'=\angle BA'L$).

It's now obvious that triangles $\triangle BO_1L$ and $\triangle BO_2B'$ have the same angles. Having in mind that $BB'=BL$ and by applying ASA:

$$\triangle BO_1L\cong\triangle BO_2B'$$

In other words, both circles have the same radius and, consequentially:

$$\angle BO_1L=\angle BO_2B',\space\angle BO_1A'=\angle BO_2A'$$

$$\implies \angle A'O_1L=\angle A'O_2B'\implies A'B'=A'L$$

• (1) The diagram in your post has jumped too many steps ahead. That is why I asked in my comment that “why $\angle K$ and $\angle L$ are ... equal”. (2) Some of the detailed explanation can be skipped because you have proved that PK and PL are tangent pairs to circle B. Just apply the properties of the tangents. (3) In proving $\triangle BA'B'\cong \triangle BA'L$, you have used “ASS” which is not an acceptable approach according to the standard Euclidean geometry. However, you can argue that the case is allowable if the triangles involved are obtuse. But note that some still resist that. – Mick Jun 26 '18 at 9:14
• @Mick: (1) You mean the diagram from Roman's original post? I agree, his diagram suggests too many things that have to be proved first. But in my solution I did not make any assumptions. (2) Yes, you are right, I could use properties of tangents but I also wanted to prove that $\angle KLP=\angle LKP=\beta$ (3) I have modified the proof and replaced ASS part with something longer and uglier but correct. Boy, it's hard to get a single upvote here :) – Oldboy Jun 26 '18 at 11:19
• +1 For your continuous effort and your work in finding the possiblilty of the congruence of those two triangles (with proof). Geometric problems always require long (and tedious) proofs and that is why they cannot arouse some's interest. – Mick Jun 26 '18 at 17:25
• Now you have your problem solved. I would like to introduce my one-step-away -from-complete problem. math.stackexchange.com/questions/2798092/… You can also ignore the lemma and some of the may-not-be necessary constructions and start anew. – Mick Jun 26 '18 at 17:50
• @Mick Sure, I'll check it. – Oldboy Jun 26 '18 at 18:17