Răspuns:
Explicație pas cu pas:
MP║AB, ⇒ΔCMP~ΔCAB. Atunci [tex]\dfrac{CM}{CA}=\dfrac{CP}{CB}[/tex] (1)
MQ║AD, ⇒ΔCMQ~ΔCAD. Atunci [tex]\dfrac{CM}{CA}= \dfrac{CQ}{CD}=\dfrac{CD-DQ}{CD}=\dfrac{CD}{CD}-\dfrac{DQ}{CD}=1- \dfrac{DQ}{CD}.[/tex] (2)
Din (1) si (2), ⇒ [tex]\dfrac{CP}{CB}=1-\dfrac{DQ}{CD},~~~deci~~\dfrac{CP}{CB}+\dfrac{DQ}{CD}=1[/tex]
