Kimberling’s point X(25)

PERSPECTOR OF ORTHIC AND TANGENTIAL TRIANGLES X(25)

Let $\triangle A_1B_1C_1$ be the orthic triangle of $\triangle ABC.$ Let $N$ be the circumcenter of $\triangle A_1B_1C_1.$ Let $\triangle A'B'C'$ be the tangencial triangle of $\triangle ABC.$ Let $O'$ be the circumcenter of $\triangle A'B'C'.$

Prove that lines $A_1A', B_1B',$ and $C_1C'$ are concurrent at point, lies on Euler line of $\triangle ABC.$

Proof

$B'C'$ and $B_1C_1$ are antiparallel to BC with respect $\angle BAC \implies B'C' ||B_1C_1.$

Similarly, $A'C' ||A_1C_1, A'B' ||A_1B_1.$

Therefore $\triangle A_1B_1C_1 \sim \triangle A'B'C' \implies$ homothetic center of $\triangle A_1B_1C_1$ and $\triangle A'B'C'$ is the point of concurrence of lines $A_1A', B_1B',$ and $C_1C'.$ Denote this point as $K.$

The points $N$ and $O'$ are the corresponding points (circumcenters) of $\triangle A_1B_1C_1$ and $\triangle A'B'C',$ so point $K$ lies on line $NO'.$

Points $N$ and $O' = X(26)$ lies on Euler line, so $K$ lies on Euler line of $\triangle ABC.$

vladimir.shelomovskii@gmail.com, vvsss

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Kimberling’s point X(25)

PERSPECTOR OF ORTHIC AND TANGENTIAL TRIANGLES X(25)