AM=PD+PE+PF
证明:
S△ABC=BC*AM/2
等边三角形中三边相等
S△ABC=PD*BC/2+PE*AC/2+PF*AB/2
=(PD+PE+PF)*BC/2
∴BC*AM/2=(PD+PE+PF)*BC/2
∴AM=PD+PE+PF
得证
AM=PD+PE+PF
证明:
S△ABC=BC*AM/2
等边三角形中三边相等
S△ABC=PD*BC/2+PE*AC/2+PF*AB/2
=(PD+PE+PF)*BC/2
∴BC*AM/2=(PD+PE+PF)*BC/2
∴AM=PD+PE+PF
得证