Question

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

Answer 1

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:
$AB^{2}= BD^{2} + AD^{2}$=> $AD^{2}=625-400=225$=> AD=15
In triunghiul BEC, dreptunghic in E aplicam Pitagora:
$BC^{2}= BE^{2} + EC^{2}$=> $BE^{2}= 40^{2}-EC^{2}$.
In triunghiul BEA, dreptunghic in E aplicam Pitagora:
$AB^{2}= BE^{2} + EA^{2}$=> $BE^{2}= 25^{2}-EA^{2}$.
Cumulate, cele doua relatii conduc catre  $40^{2}-EC^{2}= 25^{2}-EA^{2}$=>
$1600- EC^{2}=625- EA^{2}$=>$1600- EC^{2}=625- (AC-EC)^{2}$=>
$1600- EC^{2}=625- (25-EC)^{2}$=>$1600- EC^{2}=625-625+50EC- EC^{2}$=> $50EC=1600$=> EC=32=>$BE^{2}= 40^{2}-32^{2}$=>$BE^{2}= 1600-1024$=>BE=24=>$24^{2}= 25^{2}-EA^{2}$=>EA=7.