# Are lines which pass respectively through vertices $A,B,C$ and incenter, circumcenter and orthocenter of $\Delta ABC$ concurrent?

Prove that the lines through $A$ and the incenter of $\Delta ABC$, through $B$ and the circumcenter of $\Delta ABC$, and through $C$ and the orthocenter of $\Delta ABC$ are concurrent if and only if $\cos^{2} A =\cos B \cdot \cos C$.

My attempt:

By Ceva's Theorem we have that $\cfrac {AP \cdot CQ \cdot BR}{PC \cdot BQ \cdot AR}=1$,which can be rearranged in the form $\cos^2 A=\left(\cfrac {AP \cdot CQ \cdot BR}{PC \cdot BQ \cdot AC}\right)^2$, from that I've tried to solve the problem working on these two directions:

1)Show that $\left(\cfrac {AP \cdot CQ \cdot BR}{PC \cdot BQ \cdot AC}\right)^2=\cos B \cdot \cos C$

2) Since we have also that $\cos^2 A = \left( \cfrac {AR}{AC} \right)^2$, we have to show that $\cos B \cdot \cos C = \left ( \cfrac {AR}{AC} \right)^2$

I've worked most on the second line since it seems simpler and that's what I was able to do so far:

Given that $\cos B = \cfrac {BR}{RC}$ and $\cos C =\cfrac {BC^2 +AC^2-AB^2}{2AC \cdot BC}$, I have in the end (after some algebraic manipulations):

$AC^2 +CB^2-AB^2 =\cfrac {2AR ^2 \cdot BC^2}{AC \cdot BR}$ and that's where my tombstone is. I don't know how to simplify this any further, I don't know if it is even worth to simplify it given that this might be the wrong path to take...

I know that I am not coming to a solution since I am not using the fact that $BP$ passes through the circumcenter $K$ of $\Delta ABC$ and that obviously is a key point in solving the problem but I don't know how to use this information.

Edit: I've tried the following: Let $J$ be the point of intersection of lines $BP$ (this line passes through the circumenter $K$) and line $CO$ (where $O$ is the orthocentre ) ,so now what i've to do is to prove that $A,J,Q$ are collinear,i.e. I have to prove that $$\cfrac{AP \cdot CQ \cdot BJ}{BQ \cdot AC \cdot PJ}=1 \tag 1$$ . Applying the Angle Bisector Theorem to $A$ I find $\cfrac{CQ}{BQ}=\cfrac { AC}{AB}$ from which i have got the following : $$\cos^2 A = \left ( \cfrac {PJ \cdot BQ \cdot CQ}{BJ \cdot AP} \right)^2$$ Since we know that $\cos B = \cfrac{BR}{BC}$ so i have to prove $$\cos C = \left (\cfrac { PJ \cdot BQ \cdot CQ}{BJ \cdot AP} \right)^2 \cdot \cfrac {BC}{BR}$$ but so far i was unable to do that.

Any hint ,solution is appreciated.

• I think if you can prove triangle is equilateral then this is simply proved. – Archis Welankar Nov 6 '15 at 13:22
• @ArchisWelankar If the drawing is accurate, then it need not be equilateral... – rogerl Nov 6 '15 at 13:30
• yes ,it doesn't have to be equilateral. – Nameless Nov 6 '15 at 13:39
• I am just giving an option to prove they are. there are many ways like determinant form if we know three points. – Archis Welankar Nov 6 '15 at 13:57
• @ Archis Welankar I should add that the only methods i am allowed to use are purely geometrical ones. – Nameless Nov 6 '15 at 14:26

I'll be using the usual standard notation for the elements of $\triangle ABC$.

To solve this problem we need only to find the ratios ${QB\over QC},{PA\over PC}$ and ${RA\over RB}$ in terms of the elements.

We simply apply the Angle Bisector Theorem in $\triangle ABC$ to get our first ratio: $${QB\over QC}={c\over b}\tag{i}$$

Now just apply simple trigonometry in the triangles $\triangle RBC$ and $\triangle RAC$ to get the second ratio: $${RA\over RB}={b\cos A\over a\cos B}\tag{ii}$$

For the third ratio you have to do some work. Apply the Law of sines in the triangles $\triangle BPA$ and $\triangle BPC$ and obtain the relation: $\dfrac{PA}{PC}=\dfrac{\sin C}{\sin A}\cdot\dfrac{\sin \angle ABP}{\sin \angle CBP}\tag{iii}$

Then observe that $2R\cos\angle ABP=c$ and $2R\cos\angle CBP=a$. This will finally provide us the third ratio : $${PA\over PC}={\sin 2C\over \sin 2A} \tag{iv}$$

Now just apply Ceva's thoerem and you will directly obtain the result.

NOTE: You may need to use the formula $2R\sin X=x$ where $X\in\{A,B,C\}$ and $x\in\{a,b,c\}$

EDIT: Since $2R\cos\angle ABP=c$ and $2R\cos\angle CBP=a$ we have $${\sin\angle ABP\over \sin\angle CBP}={\sqrt{4R^2-c^2}\over \sqrt{4R^2-a^2}}={\cos C\over\cos A }\tag{v}$$

Apply (v) in (iii) to get (iv)

• I am not quite sure of how we got that $\cfrac{PA}{PC}=\cfrac{ \sin 2C }{ \sin 2A}$,could you elaborate a bit ? – Nameless Nov 7 '15 at 8:18
• @Nameless check out the edit. – G-man Nov 7 '15 at 14:38
• One last question:I see that $\cfrac {\sin \angle ABP }{ \sin \angle CBP}= \cfrac { \cos C }{\cos A}$ ,by noticing $\sin \angle ABP = \cos (90-ABP)= \cos C$ and same goes for $\sin \angle CBP$, but when i look at your (v) i have the impression that you derived the above fact from a different point of view,is that the case ?(That's indeed my last question,thanks again G-man for your time) – Nameless Nov 7 '15 at 17:31
• @Nameless yeah I somehow totally missed such an obvious point, the angles were complementary all the time! I applied the pythagorean identity to get to that fact when it was so glaringly obvious just by looking at the diagram! – G-man Nov 7 '15 at 18:20