Question 60
The external angle bisector of the angle $A$ of an acute-angled triangle $ABC$ meets the circumcircle of $\triangle ABC$ at point $T$. The perpendicular from the orthocenter $H$ of $\triangle ABC$ to the line $TA$ meets the line $BC$ at point $P$. The line $TP$ meets the circumcircle of $\triangle ABC$ at point $D$.
Prove that $AB^2 + DC^2 = AC^2 + BD^2$.
Prove that $AB^2 + DC^2 = AC^2 + BD^2$.