已知抛物线y²=2px(p＞0),过焦点F作直线L与抛物线交于A,B两点,作AA1⊥准线,BB1⊥准线,取AB中点M,作MM1⊥准线.于是有∠M1FB=90°.请问这是怎么证明的?另外A1M1=M1F是怎么得到的..

已知抛物线y²=2px(p＞0),过焦点F作直线L与抛物线交于A,B两点,作AA1⊥准线,BB1⊥准线,取AB中点M,作MM1⊥准线.于是有∠M1FB=90°.请问这是怎么证明的?另外A1M1=M1F是怎么得到的..
已知抛物线y²=2px(p＞0),过焦点F作直线L与抛物线交于A,B两点,作AA1⊥准线,BB1⊥准线,取AB中点M,作MM1⊥准线.于是有∠M1FB=90°.请问这是怎么证明的?另外A1M1=M1F是怎么得到的..

已知抛物线y²=2px(p＞0),过焦点F作直线L与抛物线交于A,B两点,作AA1⊥准线,BB1⊥准线,取AB中点M,作MM1⊥准线.于是有∠M1FB=90°.请问这是怎么证明的?另外A1M1=M1F是怎么得到的..
抛物线上的点到焦点的距离等于到准线的距离
AA1=AF BB1=BF
AA1B1B是直角梯形,AA1∥BB1∥MM1
∴∠A1AM1=∠M1MA
MM1=(AA1+BB1)/2=(AF+BF)/2=AB/2=AM=BM
∴∠M1AM=∠M1MA=∠A1AM1
∵ AA1=AF MA=MA ∠M1AF=∠A1AM1
∴△A1AM1≌△FAM1【边角边】
∴∠AA1M1=∠M1FA=90° A1M1=M1F
∴∠M1FB=180°-∠M1FA=90°

我不知道....................

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