Giải thích các bước giải:
b.Ta có:
$AB//CD\to \dfrac{AB}{CM}=\dfrac{EB}{EC}=\dfrac12$
$\to CM=2AB=2a\to CN=CM=2a\to BN=BC+CN=3a$
Mà $BE=\dfrac12CE\to \dfrac{BE}{CE}=\dfrac12$
$\to\dfrac{BE}{BE+CE}=\dfrac1{1+2}$
$\to \dfrac{BE}{BC}=\dfrac13$
$\to BE=\dfrac13BC=\dfrac13a$
Vì $ABCD$ là hình vuông cạnh $A\to OA=OB=OC=OD=\dfrac{a\sqrt{2}}{2}, AC=BD=a\sqrt{2}$
$\to BO\cdot BD=BE\cdot BN(=a^2)$
$\to \dfrac{BO}{BN}=\dfrac{BE}{BD}$
Mà $\widehat{OBE}=\widehat{DBN}$
$\to\Delta BOE\sim\Delta BND(c.g.c)$
c.Ta có: $CF=\dfrac12a\to FM=CM-CF=\dfrac32a, DM=DC+CM=3a$
Lại có $AD\perp CD\to AM=\sqrt{AD^2+DM^2}=a\sqrt{10}$
Mà $AB//CM\to \dfrac{HM}{HA}=\dfrac{CM}{AB}=\dfrac32$
$\to \dfrac{HM}{HM+HA}=\dfrac3{2+3}$
$\to\dfrac{HM}{AM}=\dfrac35$
$\to HM=\dfrac35AM=\dfrac{3\sqrt{10}a}{5}$
$\to MH\cdot MA=6a^2=MC\cdot MD$
$\to \dfrac{MH}{MC}=\dfrac{MD}{MA}$
Lại có $\widehat{HMC}=\widehat{AMD}$
$\to \delta MHC\sim\Delta MDA(c.g.c)$
$\to\widehat{MHC}=\widehat{MDA}=90^o$
$\to CH\perp AM$
