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 $\Delta BDM\sim\Delta BAD\to\dfrac{DM}{AD}=\dfrac{BD}{BA}$

$\to DM=DN\to đpcm$