Răspuns:
5)
Duc BD ⊥ AC si CE ⊥ AB
Caz congruenta triunghi dreptunghic ipotenuza-unghi (I.U.)
AB = AC (ipoteza)
∡A = ∡A (ipoteza) } => ΔABD ≡ ΔACE => DB ≡ CE
6)
a) atasament
b) ∡C = 90° - ∡B = 90° - 30° = 60°
c) AM = BC/2 = 10cm/2 = 5 cm
d) Cnf T30° in Δ ABC avem ca AC = BC/2 = 10cm/2 = 5 cm
Cnf T30° in ΔACD avem ca CD = AC/2 = 5cm/2 = 2,5 cm
AM mediana => CM = BC/2 = 10cm - 5cm = 5 cm
DM = CM - CD = 5cm - 5,5cm = 2, 5 cm
7)
a) atasament
b) ∡B ≡ ∡C
BC = BC } => ∡GCB = ∡HBC
∡B = ∡C } => ∡GCA ≡ ∡FBA (diferenta de unghiuri)
∡ECA ≡ ∡FBA } => ΔABF ≡ ΔACE (IU)
AB ≡ AC
c) ΔABF ≡ ΔACE => BF ≡ EC
BF ≡ EC
∡FBC ≡ ∡ECB => ΔBCF ≡ ΔCBE (LUL)
BC = BC
d)
O = BF ∩ CE => O = ortocentru (intersectia diagonalelor)
=> AO ⊥ BC
