2.如圖,我把對(duì)角線互相垂直的四邊形叫做“垂美四邊形”.
(1)性質(zhì)探究:如圖1.已知四邊形ABCD中,AC⊥BD,垂足為O,求證:AB
2+CD
2=AD
2+BC
2.
(2)解決問題:已知AB=5,BC=4,分別以△ABC的邊BC和AB向外作等腰Rt△BCQ和等腰Rt△ABP.
①如圖2,當(dāng)∠ACB=90°,連接PQ,求PQ;
②如圖3,當(dāng)∠ACB≠90°,點(diǎn)M、N分別是AC、AP中點(diǎn)連接MN.若MN=2
,則S
△ABC=
.