![](http://hiphotos.baidu.com/zhidao/pic/item/71cf3bc79f3df8dcb920ed13ce11728b47102816.jpg)
1 |
AH2 |
1 |
AB2 |
1 |
AC2 |
1 |
AD2 |
證明如下:
連接BH并延長(zhǎng)交CD于E,連接AE.∵AB,AC,AD兩兩垂直,
∴AB⊥平面ACD.又∵AE?平面ACD,∴AB⊥AE.
在Rt△ABE中,有
1 |
AH2 |
1 |
AB2 |
1 |
AE2 |
又易證CD⊥AE,
∴在Rt△ACD中,
1 |
AE2 |
1 |
AC2 |
1 |
AD2 |
將②代入①得
1 |
AH2 |
1 |
AB2 |
1 |
AC2 |
1 |
AD2 |
1 |
AD2 |
1 |
AB2 |
1 |
AC2 |
1 |
AH2 |
1 |
AB2 |
1 |
AC2 |
1 |
AD2 |
1 |
AH2 |
1 |
AB2 |
1 |
AE2 |
1 |
AE2 |
1 |
AC2 |
1 |
AD2 |
1 |
AH2 |
1 |
AB2 |
1 |
AC2 |
1 |
AD2 |