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dau coroana problema 5

Dau Coroana Problema 5 class=

Răspuns :


MN=1/2×AC
QP=1/2×AC⇔MN=QP(a)
MN ║ AC
QP ║ AC⇒MN ║QP(b)
(a),(b)⇔MNPQ paralelogram
QP ║AC
AC⊥BD⇒QP⊥BD
Dar BD ║PN( linie mijlocie)
⇔QP⊥PN
MNPQ paralelogram
QP⊥PN
⇔MNPQ dreptunghi