1. if c is CO⊥BD in o, we can get CO=FM+FN according to the above conclusion. Since the side length of a square is 3, that is, CO=(3√2)/2, FM+FN = (3 √ 2)/2 2. Connect AP, BP and CP to get triangle area Sabp+Sbcp+. S ABP=AB×τ 1, S BCP=BC×τ2, S ACP=AC×τ3, and AB=AC=BC, so ab×τ1+ab×τ2+ab×96438+0+τ 2.