Giải thích các bước giải:
a.Ta có $CA=CD\to \Delta CAD$ cân tại $C$
$\to \widehat{ACD}=180^o-2\widehat{CAD}$
$\to \dfrac12\widehat{ACD}=90^o-\widehat{CAD}=\widehat{BAD}$
$\to \widehat{BDM}=\dfrac12\widehat{ACD}=\widehat{BAD}$
Lại có $\widehat{MBD}=\widehat{ABD}$
$\to\Delta BDM\sim\Delta BAD(g.g)$
$\to\dfrac{DB}{BA}=\dfrac{BM}{BD}$
$\to BD^2=BM.BA$
b.Ta có: $\Delta ABC$ vuông tại $A, AH\perp BC\to CA^2=CH.CB$
Mà $CA=CD\to CD^2=CH.CB$
$\to\dfrac{CD}{CH}=\dfrac{CB}{CD}$
Mà $\widehat{DCH}=\widehat{DCB}$
$\to\Delta CHD\sim\Delta CDB(c.g.c)$
$\to \widehat{CDH}=\widehat{CBD}$
Ta có:
$\widehat{ADN}=\widehat{DMA}+\widehat{MAD}$
$\to\widehat{ADN}=\widehat{MBD}+\widehat{MDB}+\widehat{MAD}$
$\to\widehat{ADN}=\widehat{MBD}+\widehat{MDB}+\widehat{MDB}$
$\to\widehat{ADN}=\widehat{MBD}+2\widehat{MDB}$
$\to\widehat{ADN}=\widehat{MBD}+\widehat{ACD}$
$\to\widehat{ADN}=\widehat{ABH}-\widehat{DBH}+\widehat{ACH}-\widehat{DCB}$
$\to\widehat{ADN}=\widehat{ABH}+\widehat{ACH}-\widehat{DBH}-\widehat{DCH}$
$\to\widehat{ADN}=90^o-(\widehat{DBH}+\widehat{DCH})$
$\to\widehat{ADN}=90^o-(\widehat{HDC}+\widehat{DCH})$
$\to\widehat{ADN}=90^o-\widehat{DHB}$
$\to\widehat{ADN}=\widehat{AHD}$
Mà $\widehat{DAN}=\widehat{DAH}$
$\to\Delta ADN\sim\Delta AHD(g.g)$
$\to\dfrac{DN}{DH}=\dfrac{AD}{AH}$
$\to DN=\dfrac{AD.DH}{AH}$
Ta có $\Delta CDH\sim\Delta CBD$
$\to \dfrac{DH}{DB}=\dfrac{CD}{CB}=\dfrac{CA}{CB}=\cos\widehat{ACB}=\cos\widehat{BAH}=\dfrac{AH}{AB}$
$\to \dfrac{DH}{AH}=\dfrac{DB}{AB}$
$\to DN=\dfrac{DB}{AB}.AD=\dfrac{DM}{AD}.AD=DM$
Vì $\Delta BDM\sim\Delta BAD\to\dfrac{DM}{AD}=\dfrac{BD}{BA}$
$\to DM=DN\to đpcm$
