If the quadrilateral PEQF is a diamond, PQ is perpendicular to EF, and the intersection point m of PQ and EF is the midpoint of PQ and EF respectively.

According to the meaning of the question

(1) Assuming that point P is in AC and forms a diamond, then

EF= (BC -VL*T)

Tan α =EF/BE, so EF=tan α(BC-VL*T), ME=EF/2.

Then the distance that point P moves on the AC side is: AP=AC-ME, that is, VP*T=6-tan α(BC-VL*T)/2, and t can be obtained.

The main thing is that the moving speed of your line is 43? Or 4.3 or 4/3? Just bring it here and do the math yourself.

(2) Suppose that the point P is on BC, because BC is perpendicular to the straight line L, a diamond cannot be formed.

(3) Assuming that point P is on AB, point P should be on the right side of straight line L. ..

Then: ME still press (1), ME=tan α(BC-VL*T) /2.

At this time, the vertical line of BC is made through point P, and the vertical foot is X, then PX=ME.

Then PX/PB=sin α, so Pb = px/sinα = me/sinα = tan α (BC-VL * t)/2sinα.

And PB=(T-3-4)*5, then calculate: (T-3-4)*5=tan α(BC-VL*T) /2sin α can calculate T.