mas. unghiului ABE = mas.ungh.CBE = 60 / 2 =30 gr.
mas. ungh. ACB = 180 - (90+60)= 30 gr.
Δ BCE isoscel cu BE = EC
mas. ungh.BEC= 180 - (30+30) =120 gr.
In Δ ABC:
cf.teoremei ungh.de 30 gr :
AB =BC / 2 =10 / 2 = 5
BC² = AB² +AC²
AC²= BC² - AB²
AC²=100 - 25
AC²=75
AC= √75 = 5√3
In Δ ABE :
m.ungh. B = 30 gr.
m.ungh.A = 90 gr.
⇒m.ungh.E = 60
cf.teoremei ungh.de 30 gr:
AE = BE/2 ⇒BE = 2AE
BE=EC ⇒ 2AE = EC
dar AE + EC = AC =5√3
deci AE + 2AE =5√3
3AE = 5√3 /3 ⇒ EC= AC -AE = 5√3 - 5√3 / 3= 10 √3 / 3
PΔ BCE = BE + EC + BC = 10 √3 /3 + 10 √3/3 +10 =10(2√3 +3) / 3
