Răspuns:
Explicație pas cu pas:
MN║DC
In ΔACD aplicam Thales
[tex]\frac{AM}{MC} =\frac{AN}{ND} \\sau\\\frac{AM}{AC}= \frac{AN}{AD}[/tex]
MP║AD
In ΔACD aplicam Thales
[tex]\frac{CP}{DP}=\frac{CM}{MA}[/tex]
Obsercam ca [tex]\frac{CM}{MA} =\frac{ND}{AN} =\frac{CP}{DP}[/tex]
[tex]\frac{ND}{AN} =\frac{CP}{DP}\\\\\frac{ND}{AN+ND} =\frac{CP}{CP+DP} \\\\\frac{ND}{AD} =\frac{CP}{CD}[/tex]
Asadar
[tex]\frac{ND}{AD} +\frac{DP}{DC} =\frac{CP}{CD} +\frac{DP}{DC} =\frac{CD}{DC} =1[/tex]
