If a secant segment and a tangent segment share an endpoint outside a circle, then the product of the lengths of the secant segment and its external segment equals the square of the length of the tangent segment.
By this theorem we can draw a diagram that models the special case.
Note that two-column proof lists each statement on the left and the justification on the right. Each statement must follow logically from the steps before it. In a two-column proof, we always start by stating the given information.
Statement1)& EA is a tangent segment.
& ED is a secant segment that
& contains the center of ⊙ O.
Reason1)& Given
To continue our proof, as it was suggested we will use the Tangent Line to Circle Theorem. By this theorem we know that OA is perpendicular to EA at point A.
Statement2)& EA⊥ OA
Reason2)& Tangent Line to Circle Theorem
Let's visualize it!
Therefore, by the definition of a right angle ∠ EAO is a right angle.
Statement3)& m∠ EAO = 90^(∘)
Reason3)& Definition of a Right Angle
From here we can conclude that △ EAO is a right triangle by the definition of a right triangle.
Statement4)& △ EAO is a right triangle.
Reason4)& Definition of a Right Triangle
Now that we have a right triangle, we can use the Pythagorean Theorem.
Statement4)& x^2+r^2=(y+r)^2
Reason4)& Pythagorean Theorem
When we simplify the right-hand side of the equation, we get y^2+2yr+r^2. Thus, by the Substitution Property of Equality we will substitute y^2+2yr+r^2 for (y+r)^2.
Statement5)& x^2+r^2=y^2+2yr+r^2
Reason5)& Substitution Property of Equality
Next, using the Subtraction Property of Equality, we will subtract r^2 from each side of the equation.
Statement6)& x^2+=y^2+2yr
Reason6)& Subtraction Property of Equality
On the right-hand side we can see that y is the common factor. By the Distributive Property, we will factor it out.
Statement7)& x^2+=y(y+2r)
Reason7)& Distributive Property
Notice that x is the length of EA, y is the length of EC, and (y+2r) is the length of ED.
Finally, we will substitute EA, EC, and ED for x, y, and (y+2r), respectively. Thus, we will complete our proof.
Statement8)& EA^2+=EC* ED
Reason8)& Substitution Property of Equality
Let's summarize the above process in a two-column table.
Statement
Reason
1.
EA is a tangent segment. ED is a secant segment that contains the center of ⊙ O.
