Ta có các tứ giác BHDK,DHIC,ABDC nội tiếp
$\Rightarrow \angle KHD=\angle KBD=\angle ACD$
mà $\angle DHI+\angle ACD=180^{\circ}\Rightarrow \angle DHI+\angle KHD=180^{\circ}\Rightarrow \angle KHI=180^{\circ}\Rightarrow$ K,H,I thẳng hàng
$\angle HCD=\angle KAD\Rightarrow \Delta HCD\sim \Delta KAD\left ( G.G \right )\Rightarrow \frac{HC}{DH}=\frac{KA}{DK}$
$\angle HBD=\angle IAD\Rightarrow \Delta HBD\sim \Delta IAD\Rightarrow \frac{HB}{DH}=\frac{IA}{AD}$
$\angle BDK=\angle BHK=\angle CHI=\angle CDI\Rightarrow \Delta BDK\sim \Delta CDI\Rightarrow \frac{BK}{KD}= \frac{IC}{DI}$
Ta có $\frac{BC}{DH}=\frac{BH}{DH}+\frac{HC}{DH}= \frac{AI}{DI}+\frac{AK}{KD}=\frac{AI}{DI}+\frac{AB}{KD}+\frac{BK}{KD}=\frac{AI}{DI}+\frac{AB}{KD}+\frac{CI}{DI}=\frac{AC}{DI}+\frac{AB}{DK}$
Ta có các tứ giác BHDK,DHIC,ABDC nội tiếp
$\Rightarrow \angle KHD=\angle KBD=\angle ACD$
mà $\angle DHI+\angle ACD=180^{\circ}\Rightarrow \angle DHI+\angle KHD=180^{\circ}\Rightarrow \angle KHI=180^{\circ}\Rightarrow$ K,H,I thẳng hàng
$\angle HCD=\angle KAD\Rightarrow \Delta HCD\sim \Delta KAD\left ( G.G \right )\Rightarrow \frac{HC}{DH}=\frac{KA}{DK}$
$\angle HBD=\angle IAD\Rightarrow \Delta HBD\sim \Delta IAD\Rightarrow \frac{HB}{DH}=\frac{IA}{AD}$
$\angle BDK=\angle BHK=\angle CHI=\angle CDI\Rightarrow \Delta BDK\sim \Delta CDI\Rightarrow \frac{BK}{KD}= \frac{IC}{DI}$
Ta có $\frac{BC}{DH}=\frac{BH}{DH}+\frac{HC}{DH}= \frac{AI}{DI}+\frac{AK}{KD}=\frac{AI}{DI}+\frac{AB}{KD}+\frac{BK}{KD}=\frac{AI}{DI}+\frac{AB}{KD}+\frac{CI}{DI}=\frac{AC}{DI}+\frac{AB}{DK}$
Em thấy chứng minh K, H, I thẳng hàng chẳng để làm gì hết ạ
- frozen2501 yêu thích