1)
a²+b²+c²=ab+ac+bc
a²+b²+c²-ab-ac-bc=0 -O inmultim cu 2 =>
2a²+2b²+2c²-2ab-2bc-2ac=0
a²+b²-2ab+b²+c²-2bc+a²+c²-2ac=0
(a-b)²+(b-c)²+(a-c)²=0
a, b si c >0 pt ca sunt laturile unui Δ
trei numere pozitive adunate dau 0, daca toate sunt 0
=>a-b=0
a-c=0
b-c=0
=>a=b=c= laturile unui triunghi echilateral
Daca a=3, inaltimea:
h=a√3/2=3√3/2
Aria=h*a/2 =a²√3/4
2)a)
<B=90-<C=60°
Comparam ΔBAD cu ΔBED
<BDA=<BDE=90
BD=bisectoarea <B=> <ABD=<EBD=<B/2=30°
DB-latura comuna,
Cf. Cazului de congruenta ULU=> ΔBAD ≡ ΔBED
=> AB≡EB=> ΔABE este isoscel
b)Am demonstrat ca ΔBAD ≡ ΔBED => DE=DA
Comparam ΔEMB cu ΔMAB observam ca ED*MB=AD*MB
=> Aria ΔEMB = Aria ΔMAB
=> AM*AB=ME*EB => AM=ME
ΔEMB ≡ ΔMAB => <MAB=<MEB=90
=> ME_|_BC
c) Stim ca DM=6cm
in ΔADB: Daca <DAB=90-DBA=60=>
<MAD=<A-EAB=90-60=30°
In ΔMAD: MDA=90, avem:
sin <MAD=1/2=MD/MA=6/MA => MA=6*2=12
In ΔMAB avem:
Sin <MBA=sin 30=1/2=MA/MB=12/MB=> MB=12*2=24
In ΔCMB, <MCB=<MBC=30=>ΔCMB =isoscel si:
MC=
MB=24=>
CA=CM+MA=24+12=
36 cm
