# 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=1-x. Apply the Pythagorean theorem at triangle ASP and we get (1-x)2=x2+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 verification. If the length of the side is 1, PB=PS=1/2. Let BT=x, and ST=1-x. Apply the Pythagorean theorem at triangle PTB and we get (1/2+1-x)2=x2+1/4, then x=2/3. Therefore BT=2/3.

The third one is, of course, Haga third theorem.

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

More trisection

This way is also well known. Here we divide the diagonal, so we divide the width and the height at the same time.

 The verification is easy. Triangle APT and triangle BCT are similar, and BC=2AP, then BT=2AT. Therefore AT=AB/3.

This way was presented by KAWAHATA Fumiaki in Origami Tanteidan Newsletter issue 29.

 The verification is easy, too. Triangle ABP and Triangle CTP are similar, and CP=AP/3, therefore CT=AB/3.

The next way was presented by NOMA Masamichi in Origami Tanteidan Newsletter issue 14. The advantage of this method is to keep the center part of paper unfolded.

 Let us verify this. If the length of the side is 1, AP=BP=1/4. Let AT=x, and PT=3/4-x. Apply the Pythagorean theorem at triangle APT and we get (1/4)2+x2=(3/4-x)2, then x=1/3.

## 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 verification.

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.

The verification is complex.

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

## N-section using Haga theorems

Using Haga theorems, you can divide into not only 5, but 7, 37, or any number you like. Following table shows the relation of point P and point T. It is common to three theorems in spite of the different folds. Isn't it amazing?

 AP/AB BT/AB 1/2 2/3 1/4 2/5 3/4 6/7 1/8 2/9 3/8 6/11 5/8 10/13 7/8 14/15 ** ** n/2r 2n/(2r+n)

The verifications are the same as before.

## More n-section

 AP/AC AT/AB 1/2 1/3 1/4 1/5 3/4 3/7 1/8 1/9 3/8 3/11 5/8 5/13 7/8 7/15 ** ** n/2r n/(2r+n)

 PC/AC TC/AB 1/4 1/3 3/8 3/5 1/8 1/7 7/16 7/9 5/16 5/11 3/16 3/13 1/16 1/15 ** ** n/2r n/(2r-n)

 AP/AB TC/BC 3/4 2/3 5/8 4/5 7/8 4/7 9/16 8/9 11/16 8/11 13/16 8/13 15/16 8/15 ** ** n/2r 2r-1/n

Note: Rivised on Nov. 5, 2005. Thank you Mr. Itay Be'erli for the correction.