M , mijloc ducem linia mijlocie MM' , M'∈ BC
MM' = ( B + b ) / 2
distanta M la BC ⇒ inaltimea Δ BMC , notata cu ME , E∈ BC
si ME ⊥ BC , ME = ?
iinaltimea trapezului h = h₁ + h₂
h₁ = inaltimea in trapezul MM'CD
h₂ = inaltimea in trapezul MM'CD
aria trapez = ( B + b) · h / 2 = 180 cm²
= MM' · h = 180 cm²
aria Δ BMC = ariaΔ MM'C + aria ΔMM'C =
= MM' · h₁/ 2 + MM' · h₂ / 2 =
= ( h₁ + h₂ ) · MM' / 2
= h · MM' / 2 = 180 cm² / 2 = 90 cm²
dar aria Δ BMC = BC · ME / 2 = 90 cm²
12cm · ME = 2 · 90 cm²
ME = 180 cm² :12 cm = 15 cm
