Giải thích các bước giải:
a.Xét $\Delta AMC, \Delta BMD$ có:
$\widehat{CAM}=\widehat{MBD}=90^o$
$\widehat{CMA}=180^o-\widehat{CMD}-\widehat{DMB}=90^o-\widehat{DMB}=\widehat{MDB}$
$\to\Delta ACM\sim\Delta BMD(g.g)$
b.Ta có $AC//DB(\perp AB)$
$\to \dfrac{EA}{EB}=\dfrac{AC}{BD}$
$\to AE.BD=AC.BE$
c.Xét $\Delta HCM,\Delta HDM$ có:
$\widehat{CHM}=\widehat{DHM}(=90^o)$
$\widehat{CMH}=90^o-\widehat{HMD}=\widehat{HDM}$
$\to\Delta HCM\sim\Delta HMD(g.g)$
$\to\dfrac{HC}{HM}=\dfrac{HM}{HD}$
$\to HM^2=HD.HC$
d.Ta có $AC//DB$
$\to\dfrac{IA}{ID}=\dfrac{AC}{BD}=\dfrac{EC}{ED}$
$\to IA.DE=ID.CE$
