a) in Δ BIC
avem <BIC=180-IBC-ICB
dar < IBC=<ABC/2
<ICB=<ABC/2
deci <BIC=180-<ABC/2-<ACB/2=180-<B/2-<C/2 =(180*2-<B-<C)/2=(180+180-<B-<C)/2
dar <A=180-<B-<C
deci <BIC=(180+<A)/2= 90+<A/2
b) daca IE || BC , atunci <EIC=<ICB
dar <ECI=<ICB
=> <EIC=<ECI => triunghiul EIC este isoscel ci EI=EC
de asemenea, daca
DI || BC =>
<DIB=<IBC
dar <DBI=<IBC
=> <DBI=<DIB
triunghiul DBI este isoscel cu DB=DI
de vreme ce avem
EI=EC
DI=DB
le adunam si avem DI+IE=DE=DB+EC
c)P (ΔADE)=AD+DE+AE=AD+DI+IE+AE
am demonstrat anterior ca :
DI=DB
IE=EC
deci : P (ΔADE)=AD+DB+AE+EC
si stim ca
AB=AD+DB
AC=AE+EC
P (ΔADE)= AB+BC=25mm
Q.E.D.!