Ai desenul atasat.
Exprimam aria ΔABC isoscel (cu AB=AC) in doua moduri:
Aria ΔABC=AriaΔAMC+AriaΔAMB
[tex] \frac{BB'*AC}{2} = \frac{MN*AC}{2} + \frac{MP*AB}{2} [/tex], adica (pentru ca AB=AC):
[tex] \frac{BB'*AC}{2} = \frac{AC}{2} (MN+MP)[/tex] si daca impartim prin [tex] \frac{AC}{2} [/tex] obtinem:
BB'=MN+MP=4 cm