Ортоцентър

[Гост]

20230518_110727.jpg (355.69 KiB) Прегледано 325 пъти

[ammornil]

Screenshot 2023-05-18 093543.png (35.06 KiB) Прегледано 323 пъти

[tex]\begin{cases} \begin{cases} \angle{ACA_{1}}=180^{\circ}-135^{\circ}=45^{\circ} \\ \angle{A_{1}AC}=90^{\circ}-\angle{ACA_{1}}=45^{\circ} \end{cases} \Rightarrow AA_{1}=A_{1}C=x \Rightarrow AC=x\sqrt{2} \\ \begin{cases} \angle{BCB_{1}}=180^{\circ}-135^{\circ}=45^{\circ} \\ \angle{B_{1}BC}=90^{\circ}-\angle{BCB_{1}}=45^{\circ} \end{cases} \Rightarrow BB_{1}=B_{1}C=y \Rightarrow BC=y\sqrt{2} \end{cases} \Rightarrow[/tex]$$ BA_{1}=x+y\sqrt{2} $$
[tex]\triangle{BA_{1}H} \rightarrow \angle{A_{1}HB}=90^{\circ}-\angle{A_{1}BH}=45^{\circ} \Rightarrow A_{1}H=BA_{1}=x+y\sqrt{2}[/tex]
[tex]\triangle{AA_{1}B} \cong \triangle{CA_{1}H} \begin{cases} \text{правоъгълни} \\ AA_{1}=CA_{1}=x \\ BA_{1}=A_{1}H=x+y\sqrt{2} \end{cases} \Rightarrow[/tex] $$ CH=AB=5cm $$