ΔDBC-dreptunghic
[tex]DB= \sqrt{DC^2-BC^2}= \sqrt{15^2-(3 \sqrt{5})^2 }= \sqrt{225-45}= \sqrt{180}=6 \sqrt{5} \\ [/tex]
Coborim inaltimea trapezului din virful B,BM----inaltime
din ΔBMC
[tex]BM= \sqrt{BC^2-MC^2}= \sqrt{45-MC^2} [/tex]
Din ΔDMB
[tex]BM= \sqrt{DB^2-DM^2}= \sqrt{180-(15-MC)^2} [/tex]
Egalam cele doua relatii
[tex]45-MC^2=180-225+30MC-MC^2 \\ 30MC=90 \\ MC=3cm \\ AB=DC-2MC \\ AB=15-2*3=11cm[/tex]