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AB^2+DC^2=AC^2+BD^2, orthocenter, ext.angle bisector, circumcircle related

Source: 2011 Belarus TST 3.2

June 14, 2020
geometrycircumcircleangle bisectororthocenter

Problem Statement

The external angle bisector of the angle AA of an acute-angled triangle ABCABC meets the circumcircle of △ABC\vartriangle ABC at point TT. The perpendicular from the orthocenter HH of △ABC\vartriangle ABC to the line TATA meets the line BCBC at point PP. The line TPTP meets the circumcircce of △ABC\vartriangle ABC at point DD. Prove that AB2+DC2=AC2+BD2AB^2+DC^2=AC^2+BD^2
A. Voidelevich
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