Ducem BM perpendicular pe DC; avem BM≡AD si AB≡DM=15 si MC=5
In ΔDBC dreptunghic in B avem <DBM+<MBC=90*
In ΔBMC dreptunghic in M avem <MBC+<MCB=90*
Deci <DBM≡<MCB, atunci ΔBMC~ΔDMB si avem: [tex] \frac{BM}{DM}= \frac{MC}{BM} [/tex] ⇒ [tex] BM^{2} =DM*MC=15*5=75[/tex]⇒[tex]BM= \sqrt{75} =5 \sqrt{3} [/tex]
In ΔBMC aplicam Pitagora : [tex] BC^{2} = BM^{2} + MC^{2} =75+25=100 \\ BC=10[/tex]
AD=BM=5√3
Perimetrul=AB+BC+CD+AD=15+10+20+5√3=45+5√3=5(9+√3)
Ducem perpendiculara din A la BC si notam cu N, deci d(A;BC)=AN.
Deoarece DB perpendicular pe BC (ipoteza) rezulta AN paralel cu DB.
Ducem perpendiculara din A pe DB si notam cu P. Atunci AP este paralel cu NB si avem ANBP dreptunghi, deci AN≡PB
In ΔADB aplicam Pitagora: [tex]DB^{2} = AD^{2} + AB ^{2}=75+225=300 \\ DB= \sqrt{300} =10 \sqrt{3} [/tex]
Avem <ABP≡<DBA⇒ΔAPB~ΔDAB⇒[tex] \frac{AB}{DB}= \frac{PB}{AB} [/tex]⇒[tex]PB= \frac{AB ^{2} }{DB} = \frac{225}{10 \sqrt{3} } = \frac{225 \sqrt{3} }{30} = \frac{15 \sqrt{3} }{2} [/tex]
Dar PB≡AN⇒d(A;BC)=15√3 / 2
