$AKDI$ là tứ giác nội tiếp $\widehat{DAI}=\widehat{DKH}$$BHDK$ là tứ giác nội tiếp $\widehat{DKH}=\widehat{DBH}$
$\Rightarrow \widehat{DBH}=\widehat{DAI}$
$\Rightarrow \triangle AID\sim \triangle BHD$
$\Rightarrow \frac{AI}{BH}=\frac{ID}{HD}\Rightarrow \frac{BH}{HD}=\frac{AI}{DI}$
cmtt $\triangle CHD\sim \triangle AKD\Rightarrow \frac{CH}{HD}=\frac{AK}{KD}$
dễ dàng c/m được $\triangle BDK\sim \triangle CDI$ dựa vào tứ giác nt
$\Rightarrow \frac{BK}{IC}=\frac{DK}{DI}\Leftrightarrow \frac{BK}{KD}=\frac{IC}{DI}$
Xét $\frac{AB}{KD}+\frac{AC}{DI}=\frac{AK}{KD}-\frac{BK}{KD}+\frac{AI}{DI}+\frac{IC}{DI}$
$=\frac{AK}{KD}+\frac{AI}{DI}=\frac{BH+CH}{HD}=\frac{BC}{HD}(dpcm)$
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