Răspuns:
a) In ΔCQM, ΔCAD, MQ||DA => T.Thales => [tex]\frac{CQ}{QD}=\frac{CM}{MA}[/tex] (1)
In ΔCMP, ΔCAB, MP||AB => T.Thales => [tex]\frac{CM}{MA}=\frac{CP}{PB}[/tex] (2)
Din (1) si (2) => [tex]\frac{CP}{PB}=\frac{CQ}{QD}[/tex]
b) In ΔCQM, ΔCAD, MQ||DA => T.F.A => [tex]\frac{CQ}{DC}=\frac{QM}{DA}[/tex]
In ΔCMP, ΔCAB, MP||AB => T.F.A => [tex]\frac{CP}{CB}=\frac{MP}{AB}[/tex]
Deoarece MQ||AD si AD||BC => MQ||PC (1)
Deoarece MP||AB si MP||DC => MP||QC (2)
Din (1) si (2)=> MQCP - paralelogram => MP=QC si MQ=PC
[tex]\frac{CP}{BC}=\frac{MP}{AB} => \frac{CP}{BC}+\frac{DQ}{CD} = \frac{MP}{AB}+\frac{DQ}{CD}[/tex] deaoarece AB=CD =>
=> [tex]\frac{MP}{CD}+\frac{DQ}{CD}=\frac{DC}{DC}=1[/tex] (MP+DQ) = DC
2) AB=BC => ΔABC - isoscel
In ΔAMN, ΔABC, MN||BC => T.Thales =>[tex]\frac{AM}{BM}= \frac{AN}{NC} <=> \frac{3}{BM} = \frac{15}{8} => 15BM=24 | :(3) => 5BM=8 => BM = \frac{8}{5}[/tex]
In ΔCPN, ΔCBA, PN||BA => T.F.A => [tex]\frac{PC}{BC}=\frac{NC}{AC} <=> \frac{PC}{AB} = \frac{NC}{AC}[/tex] deoarece AB=BC
=> [tex]\frac{PC}{BM+MA}=\frac{8}{23} <=>\frac{PC}{\frac{8}{5}+3 }=\frac{8}{23} <=>\frac{PC}{\frac{23}{5} }=\frac{8}{23} <=> \frac{5PC}{23}=\frac{8}{23} => PC = \frac{8}{5} => BP = 3[/tex]
Explicație pas cu pas:
