旋转基本图形:
![](https://video.ask-data.xyz/img.php?b=https://iknow-pic.cdn.bcebos.com/2e2eb9389b504fc248816229e4dde71191ef6dc4?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)
已知:△ABC中,∠ACB=90°,AC=BC,点M、N在AB上,且∠MAN=45°
求证:MN²=AM²+BN²
证明:∵∠ACB=90°,AC=BC
∴将△ACM绕点C逆时针旋转90°可得△BCM',
∴AM=BM',CM=CM',∠ACM=∠BCM',∠A=∠CBM'=45°,
∴∠M'CM=∠BCA=90°,
又∵∠MCN=45°,
∴∠M'CN=45°=∠MCN,
又∵CN=CN,
∴△MCN≌△M'CN,
∴MN=M'N,
∵∠M'BA=∠M'BC+∠CBA=45°+45°=90°,
∴M'N²=M'B²+BN²
∴MN²=AM²+BN²