a, Gọi $I$ là giao $AC$ và $BM$
Ta có:
$AB=4a⇒CD=4a$ do $ABCD$ là hình chữ nhật
$⇒CM=\dfrac{1}{4}CD=\dfrac{1}{4}4a=a$
Ta có: $CM=a⇒CM=\dfrac{1}{2}AD=a⇒\dfrac{CM}{AD}=\dfrac{1}{2}$ do $AD=BC=2a$
Ta có:$BC=2a;CD=4a⇒\dfrac{BC}{CD}=\dfrac{2a}{4a}=\dfrac{1}{2}$
Xét $ΔBCM$ và $ΔCDA$ có:
$\dfrac{BC}{CD}=\dfrac{CM}{DA}=\dfrac{1}{2}$
$\widehat{BCM}=\widehat{CDA}=90^o$
$⇒ΔBCM$ $\sim$ $ΔCDA(c.g.c)$
$⇒\widehat{ACD}=\widehat{MBC}$
$⇒\widehat{ACD}+\widehat{BMC}=\widehat{MBC}+\widehat{BMC}$
Hay $\widehat{CIM}=90^o$
$⇒AC⊥BM$
b, Xét $ΔCIM$ có: $CM//AB$
$⇒\dfrac{CM}{AB}=\dfrac{CI}{IA}(1)=\dfrac{a}{4a}=\dfrac{1}{4}$
Xét $ΔCIM$ và $ΔBIC$ có:
$\widehat{CIM}=\widehat{BIC}=90^o$
$\widehat{IMC}=\widehat{ICB}$
$⇒ΔCIM \sim ΔBCM(g.g)$
$⇒\dfrac{IM}{IC}=\dfrac{MC}{BC}(2)=\dfrac{a}{2a}=\dfrac{1}{2}$
Từ (1) và (2)$⇒\dfrac{CI}{IA}.\dfrac{IM}{IC}=\dfrac{1}{4}.\dfrac{1}{2}=\dfrac{1}{8}$
Hay $\dfrac{IM}{IA}=\dfrac{1}{8}$
$⇒\tan\widehat{MAC}=\dfrac{1}{8}$
