连接AC交BD于O
有O为BD中点,且OC垂直BD
S△CEB=S△PEB+S△PBC,BC=BE
S△CEB=BE*OC/2=BD*BC/4,
S△PEB+S△PBC=PF*BE/2+PG*BC/2=(PG+PF)*BC/2
所以:BD*BC/4=(PG+PF)*BC/2
PF+PG=(1/2)BD
连接AC交BD于O
有O为BD中点,且OC垂直BD
S△CEB=S△PEB+S△PBC,BC=BE
S△CEB=BE*OC/2=BD*BC/4,
S△PEB+S△PBC=PF*BE/2+PG*BC/2=(PG+PF)*BC/2
所以:BD*BC/4=(PG+PF)*BC/2
PF+PG=(1/2)BD