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#1 h Sep 12, 2020, 11:51 AM • 2 Y

Y by Mango247, Mango247

Let $P$ be a point in the interior of a triangle $ABC$ . The lines $AP, BP$ and $CP$
intersect again the circumcircles of the triangles $PBC, PCA$ and $PAB$ at $D, E$ and $F$
respectively. Prove that $P$ is the orthocenter of the triangle $DEF$ if and only if $P$ is the
incenter of the triangle $ABC$ .

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#2 h Sep 12, 2020, 12:09 PM

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By inversion with respect to $P$ we get that $D^*=A^*P\cap B^*C^*$ and similarly for $E^*$ $F^*$ . From hypothesis we have $P$ being the Incenter of $D^*E^*F^*$ , and so, since $D^*E^*F^*$ is the orthic triangle of $A^*B^*C^*$ , $P$ Is the orthocenter of $A^*B^*C^*$ . Finally, this implies that $P$ is the Incenter of the original triangle $ABC$ .

This post has been edited 2 times. Last edited by cadaeibf, Sep 12, 2020, 12:12 PM

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Mahdi_Mashayekhi

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Mahdi_Mashayekhi

#3 h Jan 17, 2022, 9:35 AM

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Assume P is incenter :
∠PAF = ∠PAB + ∠BAF = ∠PAB + ∠BPF = 90 and ∠PAE = ∠PAC + ∠CAE = ∠PAC + ∠CPE = 90 so F,A,E are collinear and DP is perpendicular to EF. with same approach we can prove P is orthocenter of DEF.

Assume P is not incenter :
PA is perpendicular to EF and from last part we know it's not perpendicular at A so we have to cases:
1 - ∠PAE and ∠PAD > 90.
so ∠PBE and ∠PCD < 90 so ∠PBE + ∠PCD = ∠PBA + ∠APE + ∠PCA + ∠APD = 180 and it's contradiction.
2 - ∠PAE and ∠PAD < 90.
so ∠PBE and ∠PCD > 90 so ∠PBE + ∠PCD = ∠PBA + ∠APE + ∠PCA + ∠APD = 180 and it's contradiction.

we're Done.

This post has been edited 1 time. Last edited by Mahdi_Mashayekhi, Jan 17, 2022, 1:53 PM

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