[tex]\it \Delta ABC-isoscel,\ \ AB=AC,\ \ BC=30\ cm\\ \\ \mathcal{P}=AB+AC+BC=80 \Rightarrow AB+AC+30=80|_{-30} \Rightarrow AB+AC=50 \Rightarrow \\ \\ \Rightarrow AB+AC=25\ cm\\ \\ AD-\ \hat\imath n\breve al\c{\it t}ime \Rightarrow AD-median\breve a \Rightarrow BD=CD=30:2=15\ cm
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[tex]\it \Delta DAB-dreptunghic,\ \hat D=90^o\ \stackrel{TP}{\Longrightarrow}\ AD^2=AB^2-BD^2=25^2-15^2=\\ \\ =625-225=400=20^2 \Rightarrow AD=20\ cm[/tex]
