$\text{a) Vì ABCD là hình bình hành nên}$ \(\widehat{BAD}=\widehat{BCD}\)

\(\Rightarrow 180^0-\widehat{BAD}=180^0-\widehat{BCD}\)

\(\Rightarrow \widehat{BAH}=\widehat{BCK}\)

$\text{Xét ΔHAB và ΔKBC ta có}$

\(\left\{\begin{matrix} \widehat{BAH}=\widehat{BCK}\\ \widehat{BHA}=\widehat{BKC}=90^0\end{matrix}\right.\Rightarrow \triangle HAB\sim \triangle KCB(g.g)\)

$\text{b) Vì ΔHAB đồng dạng ΔKCB ⇒}$ \(\frac{HB}{KB}=\frac{AB}{CB}=\frac{AB}{AD}\)

\(AB\parallel CD, BK\perp CD\Rightarrow AB\perp BK\Rightarrow \widehat{ABK}=90^0\)

$\text{Ta có:}$

\(\widehat{BAD}=180^0-\widehat{BAH}=90^0+(90^0-\widehat{BAH})=90^0+\widehat{HBA}\)

\(=\widehat{ABK}+\widehat{HBA}=\widehat{HBK}\)

$\text{Xét ΔABC và ΔBHK ta có:}$

\(\left\{\begin{matrix} \widehat{BAD}=\widehat{HBK}(cmt)\\ \frac{AB}{AD}=\frac{HB}{BK}\end{matrix}\right.\Rightarrow \triangle ABD\sim \triangle HBK(c.g.c)\)

$\text{c) Theo câu a ⇒}$ \(\frac{HA}{KC}=\frac{AB}{CB}=\frac{CD}{AD}\Rightarrow AD.HA=CD.KC\)

$\text{Do đó:}$

\(DA.DH+DC.DK=DA(DA+AH)+DK(DK-CK)\)

\(=DA^2+DK^2+CD.KC-DK.CK\)

\(=BC^2+DK^2+CD.KC-DK.CK\)

\(=BK^2+CK^2+DK^2+CD.KC-DK.CK\)

\(=BD^2+(CK^2+CD.KC-DK.CK)\)

\(=BD^2+CK(CK+CD-DK)=BD^2+CK.0=BD^2\) $\text{(đpcm)}$

Giải thích các bước giải:

 a) Chứng minh $ΔHAB \sim ΔKCB$ :

+) Vì $ABCD$ là hình bình hành

$\to  \widehat{DAB} = \widehat{DCB}$

$\to 180^o - \widehat{DAB} = 180^o - \widehat{BCD}$

$\to \widehat{HAB} = \widehat{BKC}$

Xét $ΔHAB$ và $ΔKCB$ có :

$\left\{ \begin{array}{l} \widehat{AHB} = \widehat{CKB} = 90^o\\ \widehat{HAB}= \widehat{KCB} (cmt)\end{array} \right.$

$\to ΔHAB \sim ΔKCB$ $(g.g)$

b) Chứng minh $ΔABD \sim ΔBHK$ :

Ta thấy : $\widehat{DAB} = \widehat{AHB} + \widehat{HBA} = 90^o+\widehat{HBA}$

Mà : $\widehat{HBK} = \widehat{ABK} + \widehat{HBA} = 90^o+\widehat{HBA}$

$\to \widehat{DAB} = \widehat{HBK}$

Do $ΔHAB \sim ΔKCB$ $(cmt)$

$\to \dfrac{HB}{KB} = \dfrac{AB}{BC}$

$\to \dfrac{HB}{KB} = \dfrac{AB}{AD}$

Xét $ΔABD$ và $ΔBHK$ có :

$\dfrac{HB}{KB} = \dfrac{AB}{AD}$, $\widehat{DAB} = \widehat{KBH}$ $(cmt)$

$\to ΔABD \sim ΔBHK$ $(c.g.c)$

c) Chứng minh $DA.DH+DC.DK = DB^2$ :

+) Kẻ $AM ⊥ BD; CN ⊥ BD$

$\to DM = BN$

Xét $ΔDAM$ và $ΔDBH$ có:

$\widehat{D}$ chung; $\widehat{DMA} = \widehat{DHB} = 90^o$

$\to ΔDAM \sim ΔDBH$ $(g.g)$

$\to AD.DH = DM.BD$

Tương tự ta có : $ΔDNC \sim ΔDKB$ $(g.g)$

$\to DC.DK = DN.DB$

Khi đó : $AD.DH+DC.DK = BD.(DN+DM) = BD^2$ $(đpcm)$