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