a) Do ABCD là hình thoi => BC//AD hay BC//AN; CD//AB hay CD//AM; AB = BC = CD = AD = 4
Do BC//AN (cmt) => \(\widehat{C_2}=\widehat{N_1}\) (2 góc đồng vị)
CD//AM (cmt) => \(\widehat{C_1}=\widehat{M_1}\) (2 góc đồng vị)
Xét ΔBCM và ΔDNC có:
\(\widehat{C_2}=\widehat{N_1}\) (cmt)
\(\widehat{C_1}=\widehat{M_1}\) (cmt)
=> ΔBCM ~ ΔDNC (g.g)
b) Do ΔBCM ~ ΔDNC (cmt) => \(\frac{BM}{DC}=\frac{BC}{DN}\)
=> BM.DN = DC.BC = 4.4 = 16