xet ngu giac $ABCND$ co$\widehat{BAD}=\widehat{BCD}=\widehat{BND}=90^0$=>ngu giac nay noi tiep duong tron duong kinh BD tam $I$ la trung diem $BD$ va $AC$ goi $C(a;-5-2a)=>I(\frac{a-4}2;\frac{3-2a}2)$ ma IA=IN=>$(a+4)^2+(15+2a)^2=(a-14)^2+(11-2a)^2$
$=>a=\frac{76}{140}$ (so xau nen ban tu tinh nha)$=>C;I$
goi B theo hai an thuoc duong tron tam I(co mot pt )
$=>$ toa do cua M,D =>M,N.D thang hang (pt thu hai)(khong can dung BN vuong goc voi DM vi da sd o tren)
tu hai pt => toa do cua B