已知:如图,D是△ABC的边AB上一点,CN∥AB,DN交AC于点M,MA=MC.①求证:AD=CN;②若∠BAN=90度,求证:四边形ADCN是矩形.
证明:①∵CN∥AB,∴∠DAC=∠NCA,在△AMD和△CMN中,∵ ∠DAC=∠NCA
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