如图,在正方形 ABCD 中, E 、 F 是对角线 BD 上两点,且 ∠ EAF = 45 ° ,将 ΔADF 绕点 A 顺时针旋转 90 ° 后,得到 ΔABQ ,连接 EQ ,求证:
(1) EA 是 ∠ QED 的平分线;
(2) E F 2 = B E 2 + D F 2 .
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