Yet another, and pentasection 
FUSE Tomoko showed this way in her book "Unit Origami" (1983). We also trisect the right angle here. 

Here is the valification. Because AB=BC and AB=AC, triangle ABC is equilateral. So the angle ABC is 60 degrees. Angle ABE=angle CBE, then angle CBE is 30 degrees, and angle CEB 60 degrees. So BE:EC=2:1. Angle BET=angle CET, and BT:TC=BE:EC=2:1. 
How about dividing into 5? Actually, you have already done before. 

The verification is complex. Let the length of the side 1 and AR=BS=x, then PR=1/2x, SC=1x. Triangle RTP and triangle SCT are similar, and TC=2PT, then RT=SC/2=(1x)/2, TS=2PR=12x. Because RT=1TS, RT=2x. Therefore (1x)/2=2x, then x=1/5. 