[size=150][b]This activity illustrates the connections between Theorems 4, 5, 6, and 7 from section 5.2 of the UCertify [i]College Geometry[/i] text.[/b][br][br]The diagram below shows a circle of diameter 5 and a point E exterior to the circle.[br][br]If we ask how far point E is from the circle along a given line, then there are two answers: the shorter distance from E to the near side of the circle and the longer distance from E to the far side of the circle. These two distances are the lengths of segments EA and EB in the diagram.[/size][br]

[size=150]If the blue line passes through the center of the circle, then the short distance from E to the circle is 4 units and the long distance is 9 units (4 plus the diameter). These are the shortest and longest distances from E to points on the circle. As you move B around the circle the distances change. If you move B to the top of the circle then EB becomes smaller than EA. [br][br]Although the lengths of these two line segments change, the product of these two lengths remains constant.[br][br][/size][size=150]This is true for any circle and for any point E.[br][br][size=200][b]Theorem:[/b] In the diagram below, EA x EB = EC x ED.[/size][/size]

[size=150]This is actually four theorems in one. In the diagram above, you get different versions of the theorem as you move points A, C, and E. [br][br]Version 1 (two secants): If the red line and blue lines are secants of the circle intersecting at point E outside the circle, then EA x EB = EC x ED. [br][br][This is Theorem 5 from section 5.2 of the UCertify [i]College Geometry[/i] text.][br][br][/size][size=150]If you move point C so that it coincides with point D, then the red line will be tangent to the circle. [br][br]Version 2 (one secant, one tangent): In the diagram below, EA x EB = (EC)[sup]2[/sup].[br][br][This is Theorem 6 from section 5.2 of the UCertify [i]College Geometry[/i] text.][/size]

[size=150]If you also move point B so that it coincides with point A, then you get two tangent lines. [br][br]If A is the same as B and C is the same as D, then [math]EA\times EB=EC\times ED[/math] becomes [math]\left(EA\right)^2=\left(EC\right)^2[/math] which means EA equals EC.[br][br]Version 3 (two tangents): In the diagram below, EA = EC.[br][br][This is Theorem 7 from section 5.2 of the UCertify [i]College Geometry[/i] text.][br][/size]

[size=150]If you move E so that it is on the circle, then one of the line segments shrinks to a point and the length becomes zero. The length of the other segment times zero will be zero regardless of the length of the other segment. This is not too interesting, and so we will not call this a theorem. But what if we bring point E inside the circle?[/size]

[size=150]If E is inside the circle, then AB and CD will be chords of the circle.[br][br][/size][size=150]Version 4 (two chords): If AB and CD are chords of the circle with point of intersection E, then EA x EB = EC x ED.[br][br][This is Theorem 4 from section 5.2 of the UCertify [i]College Geometry[/i] text.][br][br]If you are standing at point E inside the circle and want to walk to the circle along a chosen line through E, then there are two directions you can go. Version 4 tells us that the product of those two distances from E to the circle will be constant regardless of which line you choose. If you move E outside the circle, then this same fact becomes Version 1 if the lines are secants, Version 2 if one line is tangent, or Version 3 if two lines are tangent to the circle.[br][br][/size]