Consider the circles $B_1B_2P, C_1C_2P$. By radical axis they concur at $A$. So the perpendiculars from the $A'B'C'$ to $ABC$ vice versa coincide, so the triangles are perspective. Alternatively, consider the polar of $A'$ as $\ell_{A'}$ and the polar of $A$ as $\ell_A$, inverse of $A$ be $A''$. let they intersect at $A_3$. Let the the triangles formed by $\ell_{A'}, \ell_{B'}, \ell_{C'}$ be $A_4B_4C_4$ and note it is homothetic to $ABC$. Let the orthocentre of $A_4B_4C_4$ be $H$ and consider the circles with diametres $A_4A_3$. They have radical centre $H$ clearly, and note $PA'' \cdot PA_4 = PA'' \cdot PA \cdot k = PB'' \cdot PB \cdot k = r^2 \cdot k$ etc... where $k$ is the homothety mapping $ABC \to A_4B_4C_4$, r is radius of $(P)$, so $P$ is on the radical centre of those circles, or they are co-axial with radical axis $PH$ so done.

very nice proof! Thank you, @IDMasterz

IDMasterz wrote: So the perpendiculars from the $A'B'C'$ to $ABC$ vice versa coincide, so the triangles are perspective. It is wrong.

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Ahh, I understand your supposed counter example... Read carefully before you claim something is wrong, because when "the perpendiculars coincide", I am referring to all of them meet at a point. Thank you very much my dear, aops, math-friendly, good hearted, heavenly moyen, friend.

Can someone explain the last step in the solution, cause I couldn't understand it clearelly. Tnx

I showed both $P, H$ are the radical centres of the circles, so the circles must be co-axial with radical axis $PH$.

Let $D, E, F$ be the foots of the perpendiculars from $A, B, C$ to $BC, CA, AB$ and $Q, R, S$ be the midpoints of the segments $[A_1A_2], [B_1B_2], [C_1C_2],$ respectively. Obviously $, A'\in [PQ, B'\in [PR, C'\in [PS.$ Let the midpoint of $[PA_1]$ be $M.$ Then triangles $PA'M$ and $PA_2Q$ are similar, i.e. $PA'.PQ=PM.PA_2=\frac{PA_1.PA_2}{2}.$ Similarly, we can find $PB'.PR=\frac{PB_1.PB_2}{2}$ and $PC'.PS=\frac{PC_1.PC_2}{2}.$ So, $PA'.PQ=PB'.PR=PC'.PS.$ Let $BC=a,CA=b,AB=c,PQ=x,PR=y,PS=z,$ and $PA'=\frac{k}{x},PB'=\frac{k}{y},PC'=\frac{k}{z}.$ By Ceva's Theorem, the problem is equivalent to prove $\frac{A(ABA')}{A(ACA')}.\frac{A(BCB')}{A(BAB')}.\frac{A(CAC')}{A(CBC')}=1.$ Since $A(ABP)=\frac{zc}{2}, A(BPA')=\frac{BQ.k}{2x}, A(APA')=A(QPA')=\frac{QD.k}{2x},$ we get $A(ABA')=\frac{z.c}{2}+\frac {BD.k}{2x}=\frac{zc}{2}+\frac{k.c. \cos B}{2x}=\frac{c}{2x}(zx+k\cos B).$ Similarly, we can find the other areas. So, $\frac{A(ABA')}{A(ACA')}.\frac{A(BCB')}{A(BAB')}.\frac{A(CAC')}{A(CBC')}=1.$ and the proof is completed.

IDMasterz wrote: Consider the circles $B_1B_2P, C_1C_2P$. By radical axis they concur at $A$. So the perpendiculars from the $A'B'C'$ to $ABC$ vice versa coincide, so the triangles are perspective. Can anyone give us a reference about the result of the perspectivity of two orthologic triangles with their orthologic centers coincide with ? I wonder about if it is a well known fact with also known proof. I have in mind only a reference in greek bibliography by a friend of mine Nikos Kyriazis who published a proof of this result in 1996. Than you in advance, Kostas Vittas.

Here is a strong generalization of the problem: Let $\mathcal{C}$ be a conic with center $K$ on the plane of $\triangle ABC.$ $\triangle A_0B_0C_0$ is the polar triangle of $\triangle ABC$ WRT $\mathcal{C}.$ Then any triangle $\triangle A_1B_1C_1$ homothethic to $\triangle A_0B_0C_0$ under a homothety with center $K$ is perspective with $\triangle ABC.$ Proof: Throughout the proof we use polarity WRT $\mathcal{C}.$ Let $Z \equiv KB_0 \cap AB$ and $Y \equiv KC_0 \cap AC.$ Polars of $Z,B_0,K$ concur at the line at infinite $\Longrightarrow$ polar $\tau_Z$ of $Z$ passes through the pole $C_0$ of $AB$ parallel to $AC.$ Likewise, polar $\tau_Y$ of $Y$ is the parallel to $AB$ through $B_0.$ If $U \equiv \tau_Y \cap \tau_Z,$ then $\triangle AZY$ becomes the polar triangle of $\triangle UB_0C_0$ $\Longrightarrow$ $K \equiv AU \cap ZB_0 \cap YC_0$ is the perspector of $\mathcal{C}$ WRT $\triangle UB_0C_0,$ i.e. $A,K,U$ are collinear $\Longrightarrow$ their polars $B_0C_0,YZ$ and line at infinity concur $\Longrightarrow$ $YZ \parallel B_0C_0 \ (\star).$ As $\triangle A_1B_1C_1$ varies, the series $A_1,B_1,C_1$ are similar inducing a homography $BB_1 \mapsto CC_1$ $\Longrightarrow$ $J \equiv BB_1 \cap CC_1$ is on a fixed conic $\mathcal{K}$ through $B,C.$ When $B_1 \equiv C_1 \equiv K,$ then $J \equiv K$ and when $B_1 \equiv B_0, C_1 \equiv C_0,$ then $J$ becomes the perspector $P \equiv AA_0 \cap BB_0 \cap CC_0$ of $\mathcal{C}$ WRT $\triangle ABC.$ When $B_1 \equiv Z,$ then $C_1 \equiv Y,$ due to $(\star),$ therefore $J \equiv A$ $\Longrightarrow$ all $J$ lie on unique conic $\mathcal{K}$ through $A,B,C,P,K.$ By similar reasoning, $J' \equiv AA_1 \cap BB_1$ will lie on the same conic $\mathcal{K}$ $\Longrightarrow$ $J \equiv J'$ $\Longrightarrow$ $J \equiv AA_1 \cap BB_1 \cap CC_1,$ as desired.

vittasko wrote: Can anyone give us a reference about the result of the perspectivity of two orthologic triangles with their orthologic centers coincide with ? I wonder about if it is a well known fact with also known proof. I have in mind only a reference in greek bibliography by a friend of mine Nikos Kyriazis who published a proof of this result in 1996. Than you in advance, Kostas Vittas. Dear Mr Vittas, I know what I used was a generate form of Sondat's theorem, which states that: If two triangles are mutually orthologic and perspective, then the centres of orthology and perspectivity lie on a line. Indeed, when the centres of orthology coincide, then the line always exists. Kindest Regards, Ivan

Same as anti-inequality's solution: Lemma: $\dfrac{[ABP]}{[ABA']} = \dfrac{[CBP]}{[CBC']}$ If our lemma is true, then it should be \[\dfrac{[ABA']}{[ACA']} \cdot \dfrac{[ACC']}{[BCC']} \cdot \dfrac{[BCB']}{[ABB']} = \dfrac{[ABP]}{[ACP]}\cdot \dfrac{[ACP]}{[CBP]} \cdot \dfrac{[BCP]}{[ABP]} = 1.\] So $AA'$, $BB'$, $CC'$ will be concurrent. Let's prove the lemma: Let $A_3$ be the midpoint of $A_1A_2$. So $A_3 \in PA'$. $B_3$, $C_3$ are defined similarly. Let $PA_3$ and $AB$ meet at $X$, $PC_3$ and $BC$ meet at $Y$. Let $Z$ be the foot of altitude from $C'$ to $BC$ and $W$ be the foot of altitude from $A'$ to $AB$. $A'C' \perp BP$ because $BP$ is the radical axis of circles with center $A'$ and $C'$. $P$ is the orthocenter of $\triangle BXY$. So $XY \perp BP$ and $A'C' \parallel XY$, that is $\dfrac{XP}{XA'} = \dfrac{YP}{YC'}$. $\dfrac{XP}{XA'} = \dfrac{PC_3}{A'W} = \dfrac{[ABP]}{[ABA']}$ $\dfrac{YP}{YC'} = \dfrac{PA_3}{C'Z} = \dfrac{[CBP]}{[CBC']}$ $\dfrac{[ABP]}{[ABA']} = \dfrac{[CBP]}{[CBC']}$. $\blacksquare$

Dear Mathlinkers, you can see a synthetic proof concerninfg the perspectivity when the orthology centers coincide. http://jl.ayme.pagesperso-orange.fr/ vol. 1 Le théorème de Sondat p. 1. Sincerely Jean-Louis

Trivial with orthological triangles and Dual of Sondat's Theorem. Here is a sketch of the solution: First, consider the radical axis of given circles $(P)$, $(B')$ and $(C')$. Observe that they concur at $A.$ Which means $AP\perp B'C'.$ Similarly $A'C' \perp BP$ and $B'C' \perp CP.$ So we have two orthologic triangles $\triangle ABC$ and $\triangle A'B'C'$ which have the same orthology center $P$. Then by Dual of Sondat's Theorem they have to be pespective. We're done