# High School Geometry: Chapter I: Basic Theorem

If two chords of a circle intersect anywhere. at any angle, what can be said about the segments of each cut off by the other? The data seem to be too for any conclusion, yet an important and far reaching theorem can be formulated from only this meager amount of “given”:

Theorem 2.

If two chords intersect, the product of the segments of the one equals the product of the segments of the other.

This theorem says that in Fig. 3A (attached as a jpeg picture), $PA.PB=PC.PD$ (where the dot is the symbol for algebraic multiplication). What do we mean when we talk about “the product of two line segments”? (There is a way to multiply two lines with a ruler and a compass, but we won’t discuss it here.) What the theorem means is that the product of the respective lengths of the segments are equal. Whenever we put a line segment like PA into an equation, we shall mean the length of PA. (Although we shall soon have to come to grips with the idea of a “negative length” for the present we make no distinction between the length of PA and the length of AP; they are the same positive number).

The Greek geometers took great care to enumerate the different cases of the above theorem. Today, we prefer, when possible, to treat all variants together in one compact theorem. In Fig 3A, the chords intersect inside the circle, in 3B outside the circle, and in 3C one of the chords has become a tangent. Theorem 2 holds in all three cases, and the proofs are so much alike that one proof virtually goes for all.

You may object, and say that in Fig. 3B the chords don’t intersect. But, they do when extended, and in these blogs, we shall say that one line intersects a second when it in fact only intersects the extension of the second line. This is just part of a terminology, in quite general use today, that may be slightly more free-wheeling than what you have been accustomed to. In the same vein, P divided AB internally in Fig 3A, whereas in Fig 3B, P is said to divide the chord AB externally, and we still talk about the two segments PA and PB.

To prove theorem 2, we need to construct two lines, indicated in Fig 3A. Perhaps, you should draw them also in figures 3B and 3C, and, following the proof letter for letter, see for yourself how few changes are required to complete the proofs of these diagrams. In Fig 3A, $\angle {1}= \angle {2}$ because they are inscribed in the same circular arc. And, $\angle {5} = \angle {4}$. Therefore, triangles PCA and PBD are similar, and hence have proportional sides: $\frac {PA}{PD} = \frac {PC}{PB}$ or $PA. PB = PC.PD$

In the next blog, we will discuss means and another basic theorem. Till then, aufwiedersehen,

Nalin Pithwa