如图(1),已知矩形ABCD.(1)P为矩形内一点,求证PA²+PC²=PB²+PD²..麻烦各路英雄好汉如图(1),已知矩形ABCD.(1)P为矩形内一点,求证PA²+PC²=PB²+PD²(2)P运动到AD边上,矩形ABCD外,
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![如图(1),已知矩形ABCD.(1)P为矩形内一点,求证PA²+PC²=PB²+PD²..麻烦各路英雄好汉如图(1),已知矩形ABCD.(1)P为矩形内一点,求证PA²+PC²=PB²+PD²(2)P运动到AD边上,矩形ABCD外,](/uploads/image/z/11495778-42-8.jpg?t=%E5%A6%82%E5%9B%BE%281%29%2C%E5%B7%B2%E7%9F%A5%E7%9F%A9%E5%BD%A2ABCD.%281%29P%E4%B8%BA%E7%9F%A9%E5%BD%A2%E5%86%85%E4%B8%80%E7%82%B9%2C%E6%B1%82%E8%AF%81PA%26%23178%3B%2BPC%26%23178%3B%3DPB%26%23178%3B%2BPD%26%23178%3B..%E9%BA%BB%E7%83%A6%E5%90%84%E8%B7%AF%E8%8B%B1%E9%9B%84%E5%A5%BD%E6%B1%89%E5%A6%82%E5%9B%BE%281%29%2C%E5%B7%B2%E7%9F%A5%E7%9F%A9%E5%BD%A2ABCD.%281%29P%E4%B8%BA%E7%9F%A9%E5%BD%A2%E5%86%85%E4%B8%80%E7%82%B9%2C%E6%B1%82%E8%AF%81PA%26%23178%3B%2BPC%26%23178%3B%3DPB%26%23178%3B%2BPD%26%23178%3B%EF%BC%882%EF%BC%89P%E8%BF%90%E5%8A%A8%E5%88%B0AD%E8%BE%B9%E4%B8%8A%2C%E7%9F%A9%E5%BD%A2ABCD%E5%A4%96%2C)
如图(1),已知矩形ABCD.(1)P为矩形内一点,求证PA²+PC²=PB²+PD²..麻烦各路英雄好汉如图(1),已知矩形ABCD.(1)P为矩形内一点,求证PA²+PC²=PB²+PD²(2)P运动到AD边上,矩形ABCD外,
如图(1),已知矩形ABCD.(1)P为矩形内一点,求证PA²+PC²=PB²+PD²..麻烦各路英雄好汉
如图(1),已知矩形ABCD.(1)P为矩形内一点,求证PA²+PC²=PB²+PD²(2)P运动到AD边上,矩形ABCD外,结论是否仍然成立?{图片传不上不好意思}
如图(1),已知矩形ABCD.(1)P为矩形内一点,求证PA²+PC²=PB²+PD²..麻烦各路英雄好汉如图(1),已知矩形ABCD.(1)P为矩形内一点,求证PA²+PC²=PB²+PD²(2)P运动到AD边上,矩形ABCD外,
1、如图,由P点做PF⊥BC,PH⊥AD,PE⊥AB,PG⊥CD
PA^2=AH^2+HP^2 PC^2=CF^2+PF^2
PB^2=BF^2+PF^2 PD^2=HD^2+HP^2
所以PA^2+PC^2=AH^2+HP^2+CF^2+PF^2
PB^2+PD^2=BF^2+PF^2+HD^2+HP^2
又AH^2=BF^2 CF^2=HD^2
所以PA^2+PC^2=PB^2+PD^2
2、如图,做PE⊥BC
PB^2=BE^2+PE^2 PC^2=EC^2+PE^2
所以PA^2+PC^2=PA^2+EC^2+PE^2
PB^2+PD^2=PD^2+BE^2+PE^2
又PA^2=BE^2 PD^2=EC^2
所以PA^2+PC^2=PB^2+PD^2
所以结论仍然成立.