Let ABC be an obtuse triangle with AB=AC, and let Γ be the circle that is tangent to AB at B and to AC at C. Let D be the point on Γ furthest from A such that AD is perpendicular to BC. Point E is the intersection of AB and DC, and point F lies on line AB such that BC=BF and B lies on segment AF. Finally, let P be the intersection of lines AC and DB. Show that PE=PF. geometryperpendicular bisector