Xét tam giác BHD và tam giác CKD ta có:
\(\widehat{BHD}=\widehat{CKD}\left(=90^o\right);\widehat{BCH}=\widehat{CDK}\left(d.d\right)\)
Do đó \(\Delta BHD\simeq\Delta CKD\)(g.g)
\(\Rightarrow\dfrac{BH}{CK}=\dfrac{BD}{CD}=\dfrac{HD}{KD}\)
mà \(BD=\dfrac{1}{2}CD\Rightarrow\dfrac{BD}{CD}=\dfrac{1}{2}\)
\(\Rightarrow\dfrac{BH}{CK}=\dfrac{1}{2}\Rightarrow BH=\dfrac{1}{2}CK\)(đpcm)
Chúc bạn học tốt!!!