Trường hợp [TEX]\widehat{ABC}=90^o[/TEX] thì hiển nhiên.
Nếu [TEX]\widehat{ABC}<90^o[/TEX]:
Ta có O,H nằm tròn tam giác ABC và [tex]\widehat{AOC}=2\widehat{ABC},\widehat{AHC}=90^o-\widehat{ABC}[/tex]\
Vì ABCD lồi, [tex]\widehat{DAB}=\widehat{ABC}=\widehat{BCD} \Rightarrow 360^o-3\widehat{ABC}=\widehat{ADC} < 180^o \Rightarrow \widehat{ABC} > 60^o \Rightarrow \widehat{AOC} > \widehat{AHC}[/tex] nên O nằm trong tam giác AHC.
Lại có: [tex]\frac{\sin \widehat{AHD}}{\sin \widehat{HAD}}=\frac{DA}{HD},\frac{\sin \widehat{HCD}}{\sin \widehat{CHD}}=\frac{HD}{DC},\frac{\sin \widehat{CAD}}{\sin \widehat{ACD}}=\frac{CD}{AD} \Rightarrow \frac{\sin \widehat{AHD}}{\sin \widehat{HAD}}.\frac{\sin \widehat{HCD}}{\sin \widehat{CHD}}.\frac{\sin \widehat{CAD}}{\sin \widehat{ACD}}=1[/tex]
Đặt [tex]\widehat{CAB}=x,\widehat{ABC}=y,\widehat{BCA}=z[/tex]
Khi đó [tex]\widehat{HCD}=\widehat{BCD}-\widehat{BCH}=y-(90^o-y)=2y-90^o,\widehat{CAD}=\widehat{DAB}-\widehat{CAB}=y-x[/tex]
[tex]\widehat{HAD}=y-(90^o-y)=2y-90^o,\widehat{ACD}=\widehat{BCD}-\widehat{ACB}=y-z[/tex]
[tex]\widehat{HAO}=\widehat{OAB}-\widehat{HAB}=90^o-z-(90^o-y)=y-z,\widehat{ACO}=\widehat{CAO}=90^o-z,\widehat{HCO}=\widehat{HCA}-\widehat{ACO}=90^o-x-(90^o-y)=y-x[/tex]
Suy ra [TEX]\widehat{CAD}=\widehat{HCO},\widehat{HCD}=\widehat{HAD},\widehat{ACD}=\widehat{HAO},\widehat{ACO}=\widehat{CAO}[/TEX]
[TEX]\Rightarrow \frac{\sin \widehat{AHD}}{\sin \widehat{CHD}}.\frac{\sin \widehat{CAD}}{\sin \widehat{ACD}}=1 \Rightarrow \frac{\sin \widehat{AHD}}{\sin \widehat{CHD}}.\frac{\sin \widehat{HCO}}{\sin \widehat{HAO}}.\frac{\sin \widehat{CAO}}{\sin \widehat{ACO}}=1[/TEX]
Theo định lí Céva dạng sin thì [TEX]AO,CO,HD[/TEX] đồng quy hay [TEX]H,D,O[/TEX] thẳng hàng.
Tương tự thì [TEX]\widehat{ABC} >90^o[/TEX] ta cũng có đpcm.
[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn
