Question

Exercitiul 14 (figura 7)

Answer 1

Vezi poza ...,......................

Answer 2

∡AMN = ∡DNM (unghiuri alterne interne)

∡AMN si ∡BMN sunt suplementare ⇒ ∡BMN si ∡DNM sunt suplementare

deci m(∡BMN)+m(∡DNM)=180°  si

m(∡HMN)=m(∡BMN)/2, m(∡HNM)=m(∡DNM)/2 (HM,HN bisectoare) ⇒

m(∡HMN)+m(∡HNM) = m(∡BMN)/2+m(∡DNM)/2 = (m(∡BMN)+m(∡DNM))/2=

= 180°/2 = 90°