✔ 最佳答案
1.如下圖, 取BP=a, PQ=b, QF=c
圖片參考:
https://s.yimg.com/rk/AD09210203/o/122454456.jpg
向量:AD = (2/3)AB + (1/3)AC, AP = ((b+c)/(a+b+c))AB + (a/(a+b+c))AC/2AD //AP, (b+c):a/2 = 2:1, a = b + c --- (1) 向量:AE = (1/3)AB + (2/3)AC, AQ = (c/(a+b+c))AB + ((a+b)/(a+b+c))AC/2AE //AQ, c:(a+b)/2 = 1:2, 2c = (a + b)/2 --- (2) 解(1), (2), a:b:c = 5:3:2 BP:PQ:QF = 5:3:2
2.如下圖, 過A, D分別作PB, PC平行線AA’, DD’取A’, D’中點Q, 連PQ為所求。
圖片參考:
https://s.yimg.com/rk/AD09210203/o/42487571.jpg
梯形PBA’A與PCD’D中, 面積△PMA = △BMA’, △PND = △CND’四邊形ABCD面積 = △PA’D’ 面積QA’ = QD’, 面積△PA‘Q= △PD’Q, 故四邊形ABQP面積 = DCQP 面積