Special property!

$Let\quad M_{ 1 }\quad be\quad the\quad midpoint\quad of\quad AB\quad and\quad M\quad be\quad the\quad midpoint\quad of\quad AC.\\ BAC\quad +\quad ACT\quad =\quad BMC\quad +\quad MCM_{ 2 }...eq\quad 1.\\ Since\quad BMC\quad =\quad BAC\quad +\quad ABM\quad and\quad ACT\quad -\quad MCM_{ 2 }\quad =\quad GCT,\\ From\quad eq\quad 1.,\quad BAC\quad +\quad GCT\quad =\quad BMC\quad =\quad BAC\quad +\quad ABM,\\ Hence\quad GCT\quad =\quad ABM.\\ Hence\quad BTGC\quad is\quad cyclic.\\ Hence\quad TBM\quad =\quad TCA\quad (eq\quad 2.)\quad and\quad MBC\quad =\quad CTG.\\ But\quad CTG\quad =\quad TCB\quad (Alternate\quad angles\quad between\quad parallels)\\ Hence\quad TCB\quad =\quad MBC\quad (eq\quad 3.)\quad which\quad implies,\\ TBM\quad +\quad MBC\quad =\quad TCA\quad +\quad TCB\quad (by\quad adding\quad eq\quad 2.\quad and\quad eq\quad 3.)\\ Hence\quad Triangle\quad BM_{ 2 }C\quad is\quad isoceles.\\ Hence\quad \frac { 1 }{ 2 } AB\quad =\quad AM_{ 2 }\quad =\quad BM_{ 2 }\quad =\quad M_{ 2 }C\\ Which\quad directly\quad implies\quad that\quad ACB\quad =\quad 90\quad degrees$

by descarte's theorem we could analyze that angle ACB is a right angle.

by simple angle chasing,we see,quadrilateral BTGC is cyclic with TG||BC.so BTGC is isosceles trapezium.so CG=BC.which imples a^2+b^2=c^2