If two secant segments share the same endpoint outside a circle, then the product of the lengths of one secant segment and its external segment equals the product of the lengths of the other secant segment and its external segment.
To prove this theorem we will use the following diagram.
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)& EB and ED are secant
& segments that share the same
& endpoint outside a circle.
Reason1)& Given
To continue our proof we will write a similarity relation between △ EBC and △ EDA. Therefore, to have △ EBC and △ EDA we need to draw BC and AD.
We can draw these segments by the definition of a segment.
Statement2)& Draw BC and AD.
Reason2)& Definition of a Segment
Now that we have two triangles we can compare them. Notice that both triangles share vertex E.
By the Reflexive Property of Equality, ∠ E is congruent to itself.
Statement3)& ∠ E ≅ ∠ E.
Reason3)& Reflexive Property
& of Equality
Now, let's look at ∠ EBC and ∠ EDA. Both of them intercept the same arc, AC.
By the Inscribed Angles of a Circle Theorem (Theorem 10.11), we know that if two inscribed angles of a circle intercept the same arc, then the angles are congruent. Therefore, ∠ EBC and ∠ EDA are congruent.
Statement4)& ∠ EBC ≅ ∠ EDA.
Reason4)& Inscribed Angles of
& a Circle Theorem
With this step, the two angles of △ EBC are congruent to the corresponding two angles of △ EDA. From here, by the Angle-Angle (AA) Similarity Theorem (Theorem 8.3) we can conclude that the triangles are similar.
Statement5)& △ EBC ~ △ EDA
Reason5)& AA Similarity Theorem
Since the corresponding side lengths of congruent triangles are proportional, we can write the following proportion.
Statement6)& EA/EC=ED/EB
Reason6)& Corresponding side lengths
& of congruent triangles
& are proportional.
Finally, using the Cross Product Property we will complete the proof.
Statement7)& EA * EB= EC * ED
Reason7)& Cross Product Property
Let's summarize the above process in a two-column table.
Statement
Reason
1.
EB and ED are secant segments that share the same endpoint outside a circle.
1.
Given
2.
Draw BC and AD.
2.
Definition of a Segment
3.
∠ E ≅ ∠ E
3.
Reflexive Property of Equality
4.
∠ EBC ≅ ∠ EDA
4.
Inscribed Angles of a Circle Theorem
5.
△ EBC ≅ △ EDA
5.
AA Similarity Theorem
6.
EA/EC=ED/EB
6.
Corresponding side lengths of congruent triangles are proportional.
