AD⊥BC }
MA⊥(ABC) } Teorema celor trei perpendiculare ( T3⊥) => MD⊥BC
BC ⊂ (ABC) }
MD=d(M,BC)
AD= (c1*c2) / ip = (3*4 )/5 = 12/5
MD²=1² + 144/25=169/25
MD=[tex]\frac{\sqrt{169} }{25}[/tex] = [tex]\frac{13}{5}[/tex]
Distanta de la punctul M la latura BC este 13/5
