Question

Ajutatima la ultimele exerciti 1 si 2 de jos Va rog frumos

Answer 1

Răspuns:

a) In ΔCQM, ΔCAD, MQ||DA => T.Thales => $\frac{CQ}{QD}=\frac{CM}{MA}$ (1)

In ΔCMP, ΔCAB, MP||AB => T.Thales => $\frac{CM}{MA}=\frac{CP}{PB}$ (2)

Din (1) si (2) => $\frac{CP}{PB}=\frac{CQ}{QD}$

b) In ΔCQM, ΔCAD, MQ||DA => T.F.A => $\frac{CQ}{DC}=\frac{QM}{DA}$

In ΔCMP, ΔCAB, MP||AB => T.F.A => $\frac{CP}{CB}=\frac{MP}{AB}$

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

$\frac{CP}{BC}=\frac{MP}{AB} => \frac{CP}{BC}+\frac{DQ}{CD} = \frac{MP}{AB}+\frac{DQ}{CD}$ deaoarece AB=CD =>

=> $\frac{MP}{CD}+\frac{DQ}{CD}=\frac{DC}{DC}=1$ (MP+DQ) = DC

2) AB=BC => ΔABC - isoscel

In ΔAMN, ΔABC, MN||BC => T.Thales =>$\frac{AM}{BM}= \frac{AN}{NC} <=> \frac{3}{BM} = \frac{15}{8} => 15BM=24 | :(3) => 5BM=8 => BM = \frac{8}{5}$

In ΔCPN, ΔCBA, PN||BA => T.F.A => $\frac{PC}{BC}=\frac{NC}{AC} <=> \frac{PC}{AB} = \frac{NC}{AC}$ deoarece AB=BC

=> $\frac{PC}{BM+MA}=\frac{8}{23} <=>\frac{PC}{\frac{8}{5}+3 }=\frac{8}{23} <=>\frac{PC}{\frac{23}{5} }=\frac{8}{23} <=> \frac{5PC}{23}=\frac{8}{23} => PC = \frac{8}{5} => BP = 3$

Explicație pas cu pas:

Answer 2

Răspuns:

Explicație pas cu pas: