∴PNBC=AEAD，
That is x 120 = 80? x80，
X = 48 mm;
(2) Let PN=xmm, and we can get △APN∽△ABC from the condition.
∴PNBC=AEAD，
That is x 120 = 80? PQ80，
Solution PQ = 80-23x.
∴S=PN？ PQ = x(80-23x)=-23 x2+80x =-23(x-60)2+2400，
The maximum value of ∴S is 2400mm2；;
(3) According to the area △ABC, ∫AB = 150mm, AC= 100mm,
∴ The height of side AB = 120×80 150=64mm,
AC side height = 120×80 100=96mm,
According to the method of (1), find the side length of the inscribed square of AB side, x 150 = 64? x64，
x = 4800 107≈45；
Find the side length of the inscribed square on the AC side, x 100 = 96? x96，
The solution is x = 480098≈49mm；;
(4) According to the conclusion of (1)(3), the side length of the inscribed square is the area of the triangle divided by half of the sum of the corresponding side and the height of the side.
Therefore, in order to maximize the area of the inscribed square of the triangle on the side A,
Must be a+ha < b+HB and a+ha < c+HC.