In triunghiul isoscel ABC , AB = AC , se duc inaltimile AD perpendicular pe BC , D apartine pe BC si BE perpendicular pe AC , E apartine pe AC. Stiind ca AB = 25 , BC = 40 , aflati AD , BE ,AE , CE . Mersi anticipat
AD perpendicular pe BC=> AD e mediana=> D e mijlocul lui BC=>BD=BC/2=> BD=20 si DC=20. In triunghiul ABD, dreptunghic in D aplicam Pitagora: [tex] AB^{2}= BD^{2} + AD^{2} [/tex]=> [tex] AD^{2}=625-400=225[/tex]=> AD=15 In triunghiul BEC, dreptunghic in E aplicam Pitagora: [tex] BC^{2}= BE^{2} + EC^{2} [/tex]=> [tex] BE^{2}= 40^{2}-EC^{2} [/tex]. In triunghiul BEA, dreptunghic in E aplicam Pitagora: [tex] AB^{2}= BE^{2} + EA^{2} [/tex]=> [tex] BE^{2}= 25^{2}-EA^{2} [/tex]. Cumulate, cele doua relatii conduc catre [tex] 40^{2}-EC^{2}= 25^{2}-EA^{2} [/tex]=> [tex]1600- EC^{2}=625- EA^{2} [/tex]=>[tex]1600- EC^{2}=625- (AC-EC)^{2}[/tex]=> [tex]1600- EC^{2}=625- (25-EC)^{2}[/tex]=>[tex]1600- EC^{2}=625-625+50EC- EC^{2} [/tex]=> [tex]50EC=1600[/tex]=> EC=32=>[tex] BE^{2}= 40^{2}-32^{2} [/tex]=>[tex] BE^{2}= 1600-1024[/tex]=>BE=24=>[tex] 24^{2}= 25^{2}-EA^{2} [/tex]=>EA=7.