от Nathi123 » 19 Фев 2017, 14:13
Нека [tex]BB_{1 }\bot AC\Rightarrow H\in BB_{1 }[/tex] . От [tex]\Delta BB_{1 }C\Rightarrow \angle B_{1 }BC=\angle B_{1 }CB=45^\circ ;[/tex]
От [tex]\Delta HA_{1 }B\Rightarrow \angle A_{1 }BH= \angle A_{1 }HB =45^\circ ( \angle A_{1 }BH = \angle CBB_{1 }=45^\circ; \angle HA_{1 }B=90^\circ)\Rightarrow HA_{1 }=BA_{1 }= 6 cm \Rightarrow CA_{1 }=BC - BA_{1 }=20 - 6 =14 cm.[/tex]
От [tex]\Delta CAA_{1 } \Rightarrow \angle ACA_{1 } =45^\circ[/tex] ; [tex]\angle CA_{1 }A =90^\circ \Rightarrow \angle CAA_{1 } =45^\circ \Rightarrow AA_{1 }=CA_{1 } = 14[/tex] cm.
[tex]\Rightarrow AH= AA_{1 } - HA_{1 } = 14 - 6 = 8[/tex] cm.