Fie BE⊥AD, E∈AD
si BF⊥AC, F∈AC
Arie BEDC=Arie ABCD-ArieAOB-ArieAOE
Arie ABCD=100 91)
AE||BC⇒(teoThales)ΔAOE asemenea ΔCOB
AO/OC=EO/OB=3/5⇒Aria AOB/Aria BOC= 3/5 ( aceeasi inaltime BF)
cum Arie ΔABC=Ariea ABCD:2=100:2=50⇒Arie AOB=(3/8)*50=
150/8=75/4cm² (2)
am tinut cont ca AO/AC= AO/(AO+OC)= 3/(3+5)=3/8
AO/OC=EO/OB=3/5=AO/OC
Arie AOE/ArieBOC=(3/5)²=9/25 (patratul raportului de asemanare)
dar ArieBOC=(5/8)*50=250/8=125/4
⇒Arie AOE=(9/25)*125/4=45/4 (3)
din (1) (2) si (3)⇒Arie BEDC=Arie ABCD-ArieAOB-ArieAOE=
100-75/4-45/4=100-120/4=100-30=70cm², cerinta
