in triunghiul BDC se verifica teorema lui Pitagora:
BD^2 +DC^2=BC^2 ( 20+5 = 25 ) => triunghiul BDC treptunghic in D
<BDC = 90
b) triunghiurile BDC si ADB sunt dreptunghice si asemenea
BD/BC=AB/DC=AD/BD
√5/5=AB/2√5=AD/√5 =>
AB= 2
AD=1
P = 2+1+5+2√5 = 8+2√5
C) triunghiul EAD asemenea cu EBC
EA/EB=ED/EC=AD/BC
EA/(EA+2)=ED/(ED+2√5) = 1/5
5EA=EA+2 =>4EA=2=>EA=1/2=>EB=2,5
Aria EBC=2,5*5/2=6.25