bluecarneal wrote: Let $AH_A$, $BH_B$ and $CH_C$ be the altitudes of triangle $\triangle ABC$. Prove that the triangle whose vertices are the intersection points of the altitudes of $\triangle AH_BH_C$, $\triangle BH_AH_C$ and $\triangle CH_AH_B$ is congruent to $\triangle H_AH_BH_C$. Im realy excited to see aproof of that thanks all

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mathfighter wrote: bluecarneal wrote: Let $AH_A$, $BH_B$ and $CH_C$ be the altitudes of triangle $\triangle ABC$. Prove that the triangle whose vertices are the intersection points of the altitudes of $\triangle AH_BH_C$, $\triangle BH_AH_C$ and $\triangle CH_AH_B$ is congruent to $\triangle H_AH_BH_C$. Im realy excited to see aproof of that thanks all For ease of notation, let D,E,F be the feet of the altitudes of triangle ABC Denote the orthocenters of triangles AEF, DBF, DEC by J, K, L We have to show Triangles DEF and JKL are congruent. It is well known that AH = 2 OX = 2R cos A (where X is the midpoint of BC) By analogy, in triangle AEF we have EJ = AH cos AEF = AH cos B =2R cos A Cos B Similarly DK = BH cos BDF = BH cos A = 2R cos B cos A Follows EJ = DK. Also EJ // DK Hence DEJK is a parallelogram and therefore DE = JK In the same way we show EF = KL and DF = JL Hence triangles DEF and JKL are congruent (Note that A,E,F are concyclic with H and they lie on the circle on AH as diameter) (O, H denote as usual the circum- and ortho-centers of triangle ABC)

Let $O_A$ be the orthocenter of $AH_CH_B$. Define in the same way $O_B$ and $O_C$. Since $H_CO_B \perp BC$ and $HH_A \perp BC$ we have $H_CO_B \parallel HH_A$. Similarly we have $H_AO_B \parallel HH_C$ and therefore $H_CHH_AO_B$ is a parallelogram with $O_BH_A = H_CH$ In a similar way we have that $O_AH_BHH_C$ is a parallelogram and therefore $H_CH = O_AH_B$ Hence $O_BH_A = H_CH = O_AH_B$ (1) Clearly $H_BO_A \perp AB$ and $O_BH_A \perp BC$ which implies $O_AH_B \parallel O_BH_A$. (2) From (1) and (2) we have $O_AH_BH_AO_B$ is a parallelogram and therefore $O_AO_B=H_AH_B$. In the same way we prove $O_AO_C= H_AH_C$ and $O_BO_C = H_BH_C$ which gives the desired result.