Clawson point

For a given triangle ${\displaystyle \triangle ABC}$ let ${\displaystyle \triangle H_{a}H_{b}H_{c}}$ be its orthic triangle and ${\displaystyle \triangle T_{a}T_{b}T_{c}}$ the triangle formed by the outer tangents to its three excircles. These two triangles are similar and the Clawson point is their center of similarity, therefore the three lines ${\displaystyle T_{a}H_{a},T_{b}H_{b},T_{c}H_{c}}$ connecting their vertices meet in a common point, which is the Clawson point.[2][3]

Clawson point ${\displaystyle P}$ as the perspective center of the perspective triangles ${\displaystyle \triangle ABC}$ and ${\displaystyle \triangle A'B'C'}$

For a triangle ${\displaystyle \triangle ABC}$ its circumcircle intersects each of its three excircles in two points. The three lines through those points of intersections form a triangle ${\displaystyle \triangle A^{\prime }B^{\prime }C^{\prime }}$. This triangle and the triangle ${\displaystyle \triangle ABC}$ are perspective triangles with the Clawson point being their perspective center. Hence the three lines ${\displaystyle AA^{\prime },BB^{\prime },CC^{\prime }}$ meet in the Clawson point.[1]