ΔABC dr in A
Reciproca teoremei lui Pitagora
5²=3²+4²
25=9+16
25=25
Teorema bisectoarei
[tex]\frac{AB}{AE} =\frac{BC}{EC} \\\\\frac{3}{AE} =\frac{5}{EC} =k\\\\AE=\frac{k}{3} \\\\EC=\frac{k}{5}[/tex]
[tex]AE+EC=AC=4[/tex]
[tex]\frac{k}{3}+\frac{k}{5}=4[/tex]
5k+3k=60
8k=60
[tex]k=\frac{15}{2} \\\\AE=\frac{15}{2}\times \frac{1}{3} =2,5\\\\[/tex]