A geometry problem by Rudresh Tomar

$Join\quad EH\quad and\quad DF.\quad \\ AS\quad we\quad know\quad that\quad the\quad angle\quad between\quad a\quad tangent\quad and\quad a\quad chord\quad is\\ equal\quad to\quad the\quad inscribed\quad angle on\quad the\quad opposite\quad side\quad of\quad the\quad chord\quad \\ \therefore \quad \angle GHE=\angle GEH=\angle ECH={ \quad 45 }^{ \circ }\\ similarly,\\ \angle GFD=\angle GDF=\angle DBF=\quad { 45 }^{ \circ }\\ \therefore \quad \triangle ABC\quad is\quad a\quad right-angled\quad triangle\quad with\quad AB=AC\\ \therefore \quad 2{ AB }^{ 2 }=50\\ so\quad AB=AC=5\\ so\quad AB\times AC=25$