Giải thích các bước giải:

a.Ta có : $BA=BD, CA=CD\to\Delta ABC=\Delta DBC(c.c.c)$ 

b.Từ câu a $\to \widehat{BDC}=\widehat{BAC}=90^o$

$\to \widehat{BAC}+\widehat{BDC}=90^o+90^o=180^o$

$\to ABDC$ là tứ giác nội tiếp

c.Ta có :

$\widehat{AND}=\dfrac 12\widehat{ACD}=\widehat{ACB}$
$\widehat{AMD}=\dfrac 12\widehat{ABD}=\widehat{ABC}$

$\to \widehat{AND}+\widehat{AMD}=\widehat{ABC}+\widehat{ACB}=90^o$

$\to 180^o-\widehat{MDA}-\widehat{MAD}+180^o-\widehat{ADN}-\widehat{DAN}=90^o$

$\to 360^o-(\widehat{MDA}+\widehat{ADN})-(\widehat{MAD}+\widehat{DAN})=90^o$

$\to 360^o-\widehat{MDN}-90^o=90^o$

$\to \widehat{MDN}=180^o\to M, D, N $ thẳng hàng

d.Ta có :
$\widehat{ANM}=\widehat{ACB}, \widehat{AMN}=\widehat{ABC}$ câu c

$\to\Delta ABC\sim\Delta AMN(g.g)$
$\to\dfrac{MN}{BC}=\dfrac{AM}{AB}\le \dfrac{2AB}{AB}=2$

$\to MN\le 2BC$

Dấu = xảy ra khi $ A , B, M$ thẳng hàng