Giải thích các bước giải:

Sửa đề: Tam giác $ABC$ vuông tại $A$

Bài làm:

a.Ta có $\widehat{CHA}=\widehat{CAB}=90^o,\widehat{ACH}=\widehat{ACB}$

$\to\Delta CHA\sim\Delta CAB(g.g)$

$\to\dfrac{CH}{CA}=\dfrac{CA}{CB}$

$\to CA^2=CH.CB$

Lại có $\widehat{CKA}=\widehat{CAM}=90^o,\widehat{ACK}=\widehat{ACM}$
$\to\Delta CKA\sim\Delta CAM(g.g)$

$\to\dfrac{CK}{CA}=\dfrac{CA}{CM}$

$\to CA^2=CK.CM$

$\to CK.CM=CH.CB$

$\to\dfrac{CK}{CB}=\dfrac{CH}{CM}$

Mà $\widehat{KCH}=\widehat{MCB}$

$\to\Delta CKH\sim\Delta CBM(c.g.c)$

$\to \widehat{CKH}=\widehat{CBM}=\widehat{CBA}$

b.Ta có $\widehat{MKA}=\widehat{MAC}=90^o,\widehat{AMK}=\widehat{AMC}$

$\to\Delta MAK\sim\Delta MCA(g.g)$

$\to\dfrac{MA}{MC}=\dfrac{MK}{MA}$

$\to MA^2=MK\cdot MC$

Mà $M$ là trung điểm $AB\to MB=MA$

$\to MB^2=MK\cdot MC$

$\to\dfrac{MB}{MK}=\dfrac{MC}{MB}$

Lại có $\widehat{BMK}=\widehat{BMC}$

$\to \Delta MBK\sim\Delta MCB(c.g.c)$

$\to \widehat{MBK}=\widehat{MCB}=\widehat{KCH}, \widehat{MKB}=\widehat{MBC}=\widehat{CKH}$

$\to \Delta MKB\sim\Delta HKC(g.g)$

$\to \dfrac{MK}{HK}=\dfrac{KB}{KC}$

$\to KM.KC=KB.KH$

Lại có $\widehat{MKA}=\widehat{AKC}=90^o,\widehat{MAK}=90^o-\widehat{KAC}=\widehat{KCA}$
$\to \Delta AMK\sim\Delta CAK(g.g)$

$\to\dfrac{AK}{CK}=\dfrac{MK}{AK}$
$\to KA^2=KM.KC$

$\to KA^2=KB.KH$