证明连型斗结OA,OB则在喊租猜ΔAOM和ΔBON中OA=OBOM=ONAM=BN知ΔAOM和ΔBON全等故∠AMO=∠BNO又由AM=BN则弧郑型AM=弧BN则弧AN=弧BM则∠MAC=∠NBD则在ΔMAC和ΔNBD中∠MAC=∠NBD∠AMO=∠BNO知ΔMAC和ΔNBD相似故∠ACM=∠BDN
如图,已知圆o的半径om,on交弦ab於c,d两点,且am等于bn,求证角acm等于角dbn
Simone
发布于 2023-10-17 09:50:36
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