a) \(S_{ABCD}=AH.CD=AK.BC\)
\(\Rightarrow\dfrac{AH}{AK}=\dfrac{BC}{CD}\) mà \(BC=DA\) (ABCD là hình bình hành)
\(\Rightarrow\dfrac{AH}{AK}=\dfrac{DA}{DC}\) nên \(\dfrac{AH}{DA}=\dfrac{AK}{DC}\)
b) -Ta có: \(\widehat{HAK}+\widehat{AHC}+\widehat{HCK}+\widehat{AKC}=360^0\) (tổng 4 góc trong tứ giác AHCK).
Mà \(\widehat{AHC}=\widehat{AKC}=90^0\)
\(\Rightarrow\widehat{HAK}+90^0+\widehat{HCK}+90^0=360^0\)
\(\Rightarrow\widehat{HAK}+\widehat{HCK}=180^0\)
Mà \(\widehat{HCK}+\widehat{ADC}=180^0\) (AD//BC).
\(\Rightarrow\widehat{HAK}=\widehat{ADC}\)
-Xét △HAK và △ADC có:
\(\widehat{HAK}=\widehat{ADC}\) (cmt)
\(\dfrac{AH}{DA}=\dfrac{AK}{DC}\) (cmt)
\(\Rightarrow\)△HAK ∼ △ADC (c-g-c).
\(\Rightarrow\widehat{AKH}=\widehat{ACH}\) (2 góc tương ứng).
