An interesting phenomenon happens when two chords of the same circle intersect.
Consider the applet below which shows two intersecting chords, $\overline{BC}$BC and $\overline{DE}$DE. Which two triangles in the diagram are similar? Why?
We can prove the triangles $\Delta BFD$ΔBFD and $\Delta EFC$ΔEFC in the applet are similar by angleangle similarity and the fact that the angles intercept the same arcs. Using the similar triangles, we have the following relationship between the segment lengths.
$\triangle BFD$△BFD  $\sim$~  $\triangle EFC$△EFC 
Angleangle similarity 
$\frac{DF}{CF}$DFCF  $=$=  $\frac{BF}{EF}$BFEF 
If triangles are similar, corresponding side lengths are proportional. 
$DF\times EF$DF×EF  $=$=  $CF\times BF$CF×BF 
Cross multiply 
This relationship between the chord lengths is known as the intersecting chord theorem, and we can use it to solve for unknown lengths in the chords of a circle.
If two chords intersect each other inside a circle, the products of their segment lengths are equal.
For example, in the circle below the segment lengths denoted by $a$a, $b$b, $c$c, and $d$d have the relationship $ab=cd$ab=cd.
Solve for $x$x.
Now let's explore what happens to the lengths of two secants that intersect outside the circle.
Similar triangles are also formed when two secants intersect outside the circle. Drag the points $A$A, $C$C, and $P$P around until $\overline{CP}$CP and $\overline{AP}$AP form a pair of intersecting secants. Then, use the double arrows to walk through the steps.
Why is $\triangle CPA$△CPA similar to $\triangle BPD$△BPD?
Let's explain what's happening in each step in the case where $ABDC$ABDC forms an inscribed quadrilateral:
1  $\overline{AB}$AB and $\overline{CD}$CD are secants of circle $O$O 
Given 
2  Draw $\overline{CA}$CAand $\overline{DB}$DB 
Between any two points, there exists a line. 
3  $\angle P$∠P is congruent to itself 
Reflexive property of congruence. 
4  $\angle CDB$∠CDB and $\angle BAC$∠BAC are supplementary 
Opposite angles in an inscribed quadrilateral are supplementary. 
$\angle CDB$∠CDB and $\angle BDP$∠BDP are supplementary 
Linear pairs are supplementary. 

5  $\angle BAC\cong\angle BDP$∠BAC≅∠BDP 
Congruent supplements theorem. 
Therefore, the two triangles are congruent by angleangle similarity.
Now, since we have congruent triangles, we can relate the sides in a proportion (as we did with the lengths of chords). This gives us another theorem relating the lengths of intersecting secants.
If two secant segments intersect outside the circle at a point, then the product of the measures of one secant segment and its external secant segment is equal to the product of the measures of the other secant segment and its external secant segment.
For example, in the diagram below we have the relationship
$a\left(a+b\right)=c\left(c+d\right)$a(a+b)=c(c+d)
Find the value of $x$x.
If we move the point $A$A in the applet above so that it coincides with point $B$B, then $\overline{AP}$AP becomes tangent to circle $O$O at point $A$A. secant and tangent. The same segment relationships apply to an intersecting tangentsecant pair as two secants.
If a tangent segment and a secant segment are drawn to a circle from an exterior point, then the square of the measure of the tangent segment is equal to the product of the measures of the secant segment and its external secant segment.
For example, in the diagram below we have the relationship
$a^2=b\left(b+c\right)$a2=b(b+c)
Similarity if we move point $B$B so that it coincides with point $D$D, we create two tangent segments that meet outside the circle. By algebraic manipulation, we can show that the lengths of the two tangent segments are equal. This gives us the intersecting tangents theorem discussed previously.
If two tangent segments intersect outside the circle at the same point, then they are congruent.
Find the value of $x$x.