Question

Ex 2 testul 2 fara desen. Va rog este urgent

Answer 1

MN || AC=>m(∡BNM)=m(∡BAC)

PM || AB=> m(∡CPM)=m(∡CAB)

ΔNBM≡MPC

=>m(∡BNM)=m(∡CPM)=>m(∡ANM)=m(∡APM) (1)

m(∡BMN)=m(∡PMC)

m(∡NMP)=180°-m(∡BMN)-m(∡PMC)

m(∡BMN)=m(∡AMP)

m(∡PMC)=m(∡APN)

m(NAP)=180°-m(∡AMP)-m(∡APN)=>m(NAP)=180°-m(∡BMN)-m(∡PMC)

=>m(NAP)=m(∡NMP) (2)

Din (1) si (2) =>NAPM paralelogram deoarece are unghiurile opuse congruente