Fie D∈[BC] astfel incat AD _|_ BC.
a) m(<ACD)=45* => ΔADC-dreptunghic isoscel => AD=CD.
tg(<B)=7 <=> [tex] \frac{AD}{BD}=7 \Leftrightarrow \frac{CD}{BD} =7 \Rightarrow CD=7BD. \\ \\ CD+BD=16 \sqrt{2} \Leftrightarrow 7BD+BD=16 \sqrt{2} \Leftrightarrow 8BD=16 \sqrt{2} \Rightarrow \\ BD=2 \sqrt{2}~(cm) \Rightarrow CD=14 \sqrt{2}~cm. [/tex]
[tex]T.~Pitagora~in~ \Delta ABD~(dreptunghic~in~D)~: AD^{2}+ BD^{2}= AB^{2} \Rightarrow \\ \Rightarrow AB= \sqrt{AD^2+BD^2}= \sqrt{(14 \sqrt{2})^2+(2 \sqrt{2})^2 } = \sqrt{14*2+4*2}= \\ = \sqrt{18*2}= \sqrt{36}=6~(cm). [/tex]
[tex]\Delta ADC -dreptunghic~(in~D)~isoscel~\Rightarrow AC=CD \sqrt{2}=14 \sqrt{2}* \sqrt{2}= \\ =28~(cm).[/tex]
b) [tex] A_{ABC} = \frac{AD*BC}{2}= \frac{14 \sqrt{2}*16 \sqrt{2} }{2}= 224~(cm^2).[/tex]
[tex]c)~ A_{ABC}= \frac{CE*AB}{2} \Rightarrow CE= \frac{2 A_{ABC} }{AB}= \frac{2*224}{6} = \frac{224}{3}~(cm). [/tex]
