1.若任意三点不在同一直线上,则设有A B C D四点,则有直线AB AC AD BC BD CD
2.1)AB=20,易得AC=40.所以CD=3/5AC=24
2)由原题意得:
AB=BC,AD:DC=(AB-DB):(BC+DB)=1:2,解得AB=24,所以AC=2AB=48
3.②AQ-BQ/PQ不变.证明:因为AP=BP,所以AQ-BQ/PQ=AP+PQ-(BP-PQ)/PQ=AP+PQ-(AP-PQ)/PQ=2PQ/PQ=2
4.1)化简方程mx+4=2(x+m)得(2-m)x=2(2-m),要有无数解,则m=2,所以AB=2.
2)MN=NP+MP=1/2AP+1/2BP=1/2AB,所以不变.
3)PA+PB/PC=2PB+AB/PB+BC=2PB+AB/(PB+1/2AB)=2