CE _I_ AB
Δ AEC si Δ BEC, dreptunghice, in E
a) m(BAC) = m(EAC) = 180 -(25 +90) = 65 grade
b) m(ABC) = m(EBC)= 180 -(27+90) = 63 grade
sau :
In Δ ABC :
m(ACB) = m(ACE) + m(BCE)= 25 +27 = 52 grade
m(BAC) = 65 grade (de la pct.a)
⇒m(ABC) = 180-(52+65) = 63 grade