（1）證明：
因?yàn)?是 平行四邊形,所以
∠DAB + ∠CBA = 180
AP和BP分別平分∠DAB和∠CBA
∠PAB = ∠DAB/2
∠PBA = ∠CBA/2
∠PAB + ∠PBA = (∠DAB + ∠CBA)/2 = 180/2 = 90
∠APB = 180 - (∠PAB+∠PBA) = 90
所以AP垂直PB
（2）證明：
因?yàn)?PQ || AB ,所以
∠APD = ∠PAQ
∠APQ = ∠PAD
AP平分∠DAB,∠PAQ = ∠PAD,所以
∠APD = ∠APQ
在 △APD 和 △APQ 中
∠APD = ∠APQ
AP = AP
∠PAD = ∠PAQ
所以根據(jù)角邊角定理,
△APD ≌ △APQ
對(duì)應(yīng)邊相等,則
AD = AQ
同理可證 BQ = BC
因?yàn)槭瞧叫兴倪呅?所以
BC = AD = PQ
因此 AQ = BQ = AD = 5
AB = AQ + BQ = 10
根據(jù)勾股定理,可以推出
BP^2 = AB^2 - PA^2 = 10^2 - 8^2 = 100 - 64 = 36
所以
BP = 6
因此 △APB的面積
S = (1/2) AP * PB
= (1/2)*8*6
= 24