2、如图,AB=BC=CA=AD=√3,AP⊥CD于H,CP⊥BC交AH于点P,AP=√2,则BD=.

Question

2、如图,AB=BC=CA=AD=√3,AP⊥CD于H,CP⊥BC交AH于点P,AP= √2,则BD= .

Answer 1

证明：记BD与AH交于点Q,则由AC=AD,AH⊥CD得∠ACQ=∠ADQ．
又AB=AD,故∠ADQ=∠ABQ．
从而,∠ABQ=∠ACQ．可知A、B、C、Q四点共圆．
∵∠APC=90°+∠PCH=∠BCD,∠CBQ=∠CAQ,
∴△APC∽△BCD．
∴AC•BC=AP•BD．
∴BD=AC•BC/AP=3√2/2