Giải thích các bước giải:
a.Ta có $AB=AK\to \Delta ABK$ cân tại $A\to \widehat{AKB}=\widehat{ABK}$
$CB=CH\to\Delta CBH$ cân tại $C\to \widehat{CBH}=\widehat{CHB}$
$\to \widehat{KAB}=180^o-2\widehat{ABK}=180^o-2\widehat{CBH}=\widehat{BCH}$
Mà $ABCD$ là hình bình hành
$\to \widehat{BAD}=\widehat{DCB}$
$\to \widehat{KAD}=\widehat{KAB}+\widehat{BAD}=\widehat{BCH}+\widehat{BCD}=\widehat{DCH}$
Lại có $AK=AB=CD, AD=BC=CH$
$\to \Delta AKD=\Delta CDH(c.g.c)$
$\to DK=DH$
$\to \Delta DHK$ cân tại $D$
b.Từ câu a $\to \widehat{CDH}=\widehat{AKD}$
$\to \widehat{KDH}=\widehat{ADC}-\widehat{ADK}-\widehat{HDC}$
$\to \widehat{KDH}=(180^o-\widehat{BAD})-\widehat{ADK}-\widehat{AKD}$
$\to \widehat{KDH}=180^o-\widehat{BAD}-(\widehat{ADK}+\widehat{AKD})$
$\to \widehat{KDH}=180^o-\widehat{BAD}-(180^o-\widehat{KAD})$
$\to \widehat{KDH}=180^o-\widehat{BAD}-180^o+\widehat{KAD}$
$\to \widehat{KDH}=\widehat{KAD}-\widehat{BAD}$
$\to \widehat{KDH}=\widehat{KAB}$
Mà $\Delta DKH$ cân tại $D$
$\to \widehat{AKB}=90^o-\dfrac12\widehat{KAB}=90^o-\dfrac12\widehat{KDH}=\widehat{DKH}$
$\to \Delta AKB\sim\Delta DKH(c.g.c)$
$\to \dfrac{AB}{DH}=\dfrac{KB}{KH}$
$\to AB.KH=KB.DH$
