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.
We can model this theorem as follows.
Note that two-column proof lists each statement on the left and the justification is 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.
Reason1)& Given
To continue our proof we will need two segments that will help us to create △ EAC and △ EDA.
These segments can be drawn according to the definition of a segment.
Statement2)& Draw AC and AD.
Reason2)& Definition of a Segment
From here we will try to write a similarity relation between △ EAC and △ EDA. Notice that ∠ EDA is an inscribed angle whose intercepted arc is AC. Therefore, by the Measure of an Inscribed Angle Theorem (Theorem 10.10), the measure of ∠ EDA is half the measure of AC.
Statement3)& m∠ EDA = 1/2mAC
Reason3)& Measure of an Inscribed
& Angle Theorem
As we can see, AC is also the intercepted arc of ∠ EAC.
By the Tangent and Intersected Chord Theorem (Theorem 10.14),the measure of ∠ EAC is also the half the measure of AC.
Statement4)& m∠ EAC = 1/2mAC
Reason4)& Tangent and Intersected
& Chord Theorem
Since both ∠ EDA and ∠ EAC have the same measure of angle, we can say that by the Transitive Property of Equality they are equal.
Statement5)& m∠ EDA = m∠ EAC
Reason5)& Transitive Property
& of Equality
We also know that if two angles have the same angle measure then they are congruent.
Statement6)& ∠ EDA ≅ ∠ EAC
Reason6)& Definition of Congruent Angles
Additionally, notice that both triangles share the same vertex E.
Thus, by the Reflexive Property of Equality ∠ E is congruent to itself.
Statement7)& ∠ E ≅ ∠ E
Reason7)& Reflexive Property
& of Equality
Combining all of this information, we have two angles of △ EDA that are congruent to the corresponding angles of △ EAC. Therefore, by the Angle-Angle (AA) Similarity Theorem they are similar.
Statement8)& △ EDA ~ △ EAC
Reason8)& AA Similarity Theorem
Remember that the corresponding side lengths of similar triangles are proportional.
Statement9)& EA/ED=EC/EA
Reason9)& Definition of Similar Triangles
Finally, using the Cross Product Property we can complete our proof.
Statement10)& EA^2=EC* ED
Reason10)& Cross Product Property
Let's summarize the above process in a two-column table.
