Giải thích các bước giải:

a.Ta có $\Delta ABC$ vuông tại $A$

$\to BC^2=AB^2+AC^2=2500$

$\to BC=50$

b.Xét $\Delta BAK,\Delta BAC$ có:

chung $\hat B$

$\dfrac{BK}{BA}=\dfrac{BA}{BC}(=\dfrac35)$
$\to \Delta BAK\sim\Delta BCA(c.g.c)$

c.Từ câu b $\to \widehat{BKA}=\widehat{BAC}=90^o\to AK\perp BC$

Xét $\Delta AKC,\Delta ABC$ có: 

Chung $\hat C$

$\widehat{CKA}=\widehat{CAB}(=90^o)$

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

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

$\to CA^2=CK.CB$

d.Xét $\Delta KAB,\Delta KAC$ có:

$\widehat{AKB}=\widehat{AKC}(=90^o)$
$\widehat{KAB}=90^o-\widehat{KAC}=\widehat{KCA}$

$\to\Delta KAB\sim\Delta KCA(g.g)$

$\to \dfrac{KA}{KC}=\dfrac{KB}{KA}$

$\to \dfrac{KA}{3KE}=\dfrac{KB}{3KM}$

$\to \dfrac{KA}{KE}=\dfrac{KB}{KM}$

$\to \dfrac{KA}{KB}=\dfrac{KE}{KM}$

Mà $\widehat{BKM}=\widehat{AKE}(=90^o)$

$\to \Delta KMB\sim\Delta KEA(c.g.c)$

$\to \widehat{KAE}=\widehat{KBM}=90^o-\widehat{BMK}=90^o-\widehat{AMQ}=\widehat{QAM}$

$\to A, Q, E$ thẳng hàng