## Tuesday, December 16, 2008

### Archimedes' Book of Lemmas, Proposition #9

Problem 648: Perpendicular chords and arcs
Exercise your brain. Archimedes wrote the "Book of Lemmas" more than 2200 years ago. Solve the proposition #9 (high school level) and lift up your geometry skills. gogeometry.com/ArchBooLem09.htm

1. Here we go:
The statement to prove is equivalent to the statement that arc AD+arc CB is the length of half the circumference of the circle. We will prove the last one.

Draw the line l, which passes through the center of the circle and is parallel to AB. Now consider the reflection of the setting in l. The line of the circle is mapped to itself and now the image of A is named A', the image of B is B' etc. Now, as l is the perpendicular bisector of CD, we have D'=C and C'=D.

As AA' is perpendicular to the mirror l, it must also be perpendicular to its parallel AB. But when A'AB is a right angle on the circle line, its respective chord A'B has to be a diameter by the converse of Thales' theorem.

So, we have, that
0.5 diameter=arc A'B=arc A'D'+arc CB= arc AD+arc CB, as claimed. (by Thomas)

2. http://img841.imageshack.us/img841/5839/problem648.png
Connect AC and BD
Note that angle(APC)=angle(PDB)+angle(PBD)= 90
and angle(DPB)=angle(ACP)+angle(CAP)=90
Peter Tran

3. Let O be the center of the circle
∠APC = 90°
∴ ∠ACP + ∠CAP = 90°
Same as ∠ACD + ∠CAB = 90°
∴ ∠AOD + ∠COB = 180°
+ measure of arc BC = 180°
Similarly the other sum = 180°

4. Construction=join AC
1.ang. CAB=1/2 arc BC.......(inscribed ang.thm(I.A.t))
3.hence,
ang. CAB + ang. ACD =m 1/2(arc BC+ arc AD)....(1+2)
4.in tri ACP,
ang. CAB + ang. ACD =ang CPB....(ext ang of triangle)
5.hence,
ang CPB = m 1/2(arc BC+ arc AD)....(from 4,5)