As shown in the figure, in the RT triangle ABC, ∠ C = 90, AC = 6cm, BC = 8cm, points E and f start from two points a and B at the same time and move to points c and B along the direction of AC and Ba respectively The velocity of point E is 2cm / s, and that of point F is 1cm / s. if one of the points reaches the position, both points stop moving (1) Q. after a few seconds, the area of the triangle AEF is 16 / 5 (2) After a few seconds, EF bisects the perimeter of the RT triangle (3) Is there a line EF that bisects the perimeter and area of rtabc at the same time? If so, calculate the length of AE. If not, explain the reason

As shown in the figure, in the RT triangle ABC, ∠ C = 90, AC = 6cm, BC = 8cm, points E and f start from two points a and B at the same time and move to points c and B along the direction of AC and Ba respectively The velocity of point E is 2cm / s, and that of point F is 1cm / s. if one of the points reaches the position, both points stop moving (1) Q. after a few seconds, the area of the triangle AEF is 16 / 5 (2) After a few seconds, EF bisects the perimeter of the RT triangle (3) Is there a line EF that bisects the perimeter and area of rtabc at the same time? If so, calculate the length of AE. If not, explain the reason

1、 Make FG ⊥ AC through f point;
AC = 6, BC = 8, ∠ C = 90 ° and ab = 10;
The results show that Ag / AC = (ab-bf) / AB, that is, Ag = 0.6 (10-bf);
FG / BC = (ab-bf) / AB, that is, FG = 0.8 (10-bf);
S△AEF=AG*FG/2=0.24(10-BF)^2;（1）
When s △ AEF = 16 / 5, substitute (1) to get BF = 10-2 √ 30 / 3, or 10 + 2 √ 30 / 3 (rounding off);
Since the velocity of BG is 1cm / s, s △ AEF is 16 / 5 after (10-2 √ 30 / 3) seconds (about 6.35 seconds);
2、 Half of the circumference of RT △ ABC = (AB + AC + BC) / 2 = (10 + 6 + 8) / 2 = 12 > ab;
So point E is on the extension line of AC; it is also called FG ⊥ AC;
AE=2BF;
EG=AE-AG=2BF-0.6(10-BF)=2.6BF-6;
EF^2=EG^2+FG^2=(2.6BF-6)^2+(0.8(10-BF))^2;(2)
When EF = 12 (half of the circumference of RT △ ABC), substitute (2)
BF = (√ 809.6 + 22) / 7.4, about 6.818cm;
After 6.818 seconds, EF bisects the perimeter of RT △ ABC;
3、 RT △ ABC perimeter = 24; SRT △ ABC = 6 * 8 / 2 = 24;
Therefore, if EF bisects the cycle length, it also bisects the area;
At this time, AE = 2BF = (√ 809.6 + 22) / 3.7, about 12.636cm;