toan thi vao 10 kho

U

utan1995

Co le minh se chung minh tu giac MADB la tu giac noi tiep => goc BMD = goc BAD
=> BMD = 60*
 
J

jet_nguyen

△PBC∼△CDN(g.g) \triangle PBC \sim \triangle CDN (g.g)

=> BPa=aND=>BPBD=BDND\frac{BP}{a} = \frac{a}{ND} => \frac{BP}{BD}=\frac{BD}{ND} (a là dộ dài cạnh hình thoi)

=> △BPD∼△BDN(c.g.c)\triangle BPD \sim \triangle BDN (c.g.c)

=> MBP^+BMD^=BPD^=NBD^=MBP^+ABD^ \hat{MBP}+\hat{BMD} = \hat{BPD} = \hat{NBD} = \hat{MBP}+\hat{ABD}

=> BMD^=ABD^=600\hat{BMD} = \hat{ABD} = {60}^{0}
 
V

venus01

.............................................................................................................................................................................................
 
Last edited by a moderator:
You must log in or register to reply here.
Top Bottom