[tex]P_{ABC}=AB+BC+AC=80 <=>AB+AB+30=80<=> \\ <=>2AB+30=80=>2AB=50 => AB=AC=25(cm). [/tex]
AD-inaltime => D-mijlocul lui [BC] => BD=DC=[tex] \frac{BC}{2}= \frac{30}{2}=15(cm). [/tex]
Aplic teorema lui Pitagora in ΔABD-dreptunghic in D:
[tex] AD^{2}+ BD^{2}= AB^{2} => AD^{2}= AB^{2}- BD^{2}=> AD= \sqrt{ AB^{2}- BD^{2} } [/tex]
[tex]AD= \sqrt{ 25^{2}- 15^{2} }= \sqrt{625-225} = \sqrt{400} =20(cm). [/tex]
