Question

这道题做到这不会了，谁能给我讲讲？

Answer 1

为表述简便, 我们设:
AD = L, FH = L1, EG = L2
BD = a1, DC = a2
AF = x, AE = y
HD = x', DG = y'

三角形 HEG 中, P 是HE 和 AD 交点,
PD/L2 = x'/(x'+y')
其中, x'/x = a1/c, L2/L = (b-y)/b
PD = a1x/c * (b-y)/b * L/(x'+y')

三角形 GFH 中, P' 是GF 和 AD 交点,
P'D/L1 = y'/(x'+y')
其中, y'/y = a2/b, L1/L = (c-x)/c
P'D = a2y/b * (c-x)/c * L/(x'+y')
PD/P'D = a1/a2 * x/y * (b-y)/(c-x) = BD/DC * AF/AE * EC/BF

如楼主所述, BD/DC * AF/AE * EC/BF = 1
PD = P'D
因此, P 和 P' 重合, EH, FG, AD 交于一点 P