Haga theorems 
First, Haga first theorem is famous. 

Let us verify it. If the length of the side is 1, AP=BP=1/2. Let AS=x, and SP=1x. Apply the Pythagorean theorem at triangle ASP and we get (1x)^{2}=x^{2}+1/4, then x=3/8. Triangle ASP and triangle BPT are similar, and AP=BP=1/2, AS=3/8, therefore BT=2/3. 
Second, there is Haga second theorem. 

And here is the varification. If the length of the side is 1, PB=PS=1/2. Let BT=x, and ST=1x. Apply the Pythagorean theorem at triangle PTB and we get (1/2+1x)^{2}=x^{2}+1/4, then x=2/3. Therefore BT=2/3. 
The third one is, of course, Haga third theorem. 

The varification is rather complex. If the length of the side is 1, PB=1/2. Let BT=x, and TC=1x. Since triangle PBT and triangle TCS are similar, CS=2x(1x). Therefore ST=12x(1x). Apply the Pythagorean theorem at triangle TCS and we get (12x(1x))^{2}=(1x)^{2}+(2x(1x))^{2}. We have two solutions: x=0 or x=2/3. Since BT is not 0, BT=2/3. 