от Nathi123 » 10 Мар 2021, 14:04
Построяваме СС'||BD [tex]\Rightarrow DC||AB\Rightarrow DBC'C[/tex] - e успоредник.[tex]\Rightarrow BC'=CD = 1 AC'=AB+BC'= 9+1=10[/tex] За [tex]\triangle AC'C[/tex]
[tex]p=\frac{1}{2}(10+6+8) =12 ( BD=CC'=6 ; AC=8 )\Rightarrow S_{\triangle AC'C}=\sqrt{12.(12-8).(12-6).(12-10)}=\sqrt{12.4.6.2}=24; CH\bot AB\Rightarrow[/tex]
[tex]S_{\triangle AC'C}=\frac{1}{2}.CH.10\Rightarrow CH=\frac{24}{5}\Rightarrow S_{ABCD}=\frac{1}{2}.CH.(AB+CD)=24[/tex] т.е. [tex]S_{\triangle AC'C}= S_{ABCD}[/tex]
[tex]S_{ABCD}=\frac{1}{2}AC.BCsin\varphi\Rightarrow sin\varphi =1\Leftrightarrow \varphi =90^\circ[/tex]