Lời giải:
a)
Vì $ABCD$ là hình bình hành nên \(\widehat{BAD}=\widehat{BCD}\)
\(\Rightarrow 180^0-\widehat{BAD}=180^0-\widehat{BCD}\)
\(\Rightarrow \widehat{BAH}=\widehat{BCK}\)
Xét tam giác $HAB$ và $KCB$ ta có:
\(\left\{\begin{matrix} \widehat{BAH}=\widehat{BCK}\\ \widehat{BHA}=\widehat{BKC}=90^0\end{matrix}\right.\Rightarrow \triangle HAB\sim \triangle KCB(g.g)\)
b)
Từ hai tam giác đồng dạng phần a suy ra \(\frac{HB}{KB}=\frac{AB}{CB}=\frac{AB}{AD}\)
\(AB\parallel CD, BK\perp CD\Rightarrow AB\perp BK\Rightarrow \widehat{ABK}=90^0\)
Ta có:
\(\widehat{BAD}=180^0-\widehat{BAH}=90^0+(90^0-\widehat{BAH})=90^0+\widehat{HBA}\)
\(=\widehat{ABK}+\widehat{HBA}=\widehat{HBK}\)
Xét tam giác $ABD$ và $BHK$ có:
\(\left\{\begin{matrix} \widehat{BAD}=\widehat{HBK}(cmt)\\ \frac{AB}{AD}=\frac{HB}{BK}\end{matrix}\right.\Rightarrow \triangle ABD\sim \triangle HBK(c.g.c)\)
c)
Theo phần a suy ra \(\frac{HA}{KC}=\frac{AB}{CB}=\frac{CD}{AD}\Rightarrow AD.HA=CD.KC\)
Do đó:
\(DA.DH+DC.DK=DA(DA+AH)+DK(DK-CK)\)
\(=DA^2+DK^2+CD.KC-DK.CK\)
\(=BC^2+DK^2+CD.KC-DK.CK\)
\(=BK^2+CK^2+DK^2+CD.KC-DK.CK\)
\(=BD^2+(CK^2+CD.KC-DK.CK)\)
\(=BD^2+CK(CK+CD-DK)=BD^2+CK.0=BD^2\)
Ta có đpcm.
