Giải thích các bước giải:

a.Xét $\Delta AMB,\Delta CKB$ có:

Chung $\hat B$

$\widehat{AMB}=\widehat{CKB}(=90^o)$

$\to \Delta AMB\sim\Delta CKB(g.g)$

$\to \dfrac{AM}{CK}=\dfrac{MB}{KB}$

$\to AM.KB=MB.CK$

b.Xét $\Delta AHK,\Delta CMH$ có:

$\hat K=\hat M(=90^o)$

$\widehat{AHK}=\widehat{CHM}$

$\to \Delta KAH\sim\Delta MCH(g.g)$

$\to \dfrac{AK}{CM}=\dfrac{KH}{MH}$

$\to AK.MH=KH.MC$

c.Gọi $BH\cap AC=D$

Ta có: $AM\perp BC, CK\perp AB, CK\cap AM=H$

$\to H$ là trực tâm $\Delta ABC$

$\to BH\perp AC=D$

Xét $\Delta ADH,\Delta AMC$ có:

Chung $\hat A$

$\hat D=\hat M(=90^o)$

$\to \Delta ADH\sim\Delta AMC(g.g)$

$\to \dfrac{AD}{AM}=\dfrac{AH}{AC}$

$\to AH.AM=AD.AC$

Tương tự: $CH.CK=CD.CA$

$\to AH.AM+CH.CK=AD.AC+CD.AC=AC^2$