In Figure 1, chords QS and RT intersect at *P*. By drawing QT and RS, it can be proven that Δ QPT ∼ Δ RPS. Because the ratios of corresponding sides of similar triangles are equal, *a*/ *c* = *d*/ *b*. The *Cross Products Property* produces ( *a*) ( *b*) = ( *c*) ( *d*). This is stated as a theorem.

**Figure 1 ***Two chords intersecting inside a circle.*

*Theorem 83:* If two chords intersect inside a circle, then the product of the segments of one chord equals the product of the segments of the other chord.

**Example 1:** Find *x* in each of the following figures in Figure 2.

**Figure 2 **Two chords intersecting inside a circle.

In Figure 3, secant segments AB and CD intersect outside the circle at *E*. By drawing BC and AO, it can be proven that Δ *EBC* ∼ Δ *EDA*. This makes

**Figure 3 **Two secant segments intersecting outside a circle.

By using the *Cross‐Products Property*,

This is stated as a theorem.

*Theorem 84:* If two secant segments intersect outside a circle, then the product of the secant segment with its external portion equals the product of the other secant segment with its external portion.

**Example 2:** Find *x* in each of the following figures in 4.

**Figure 4 **More secant segments intersecting outside a circle.

In Figure 5, tangent segment AB and secant segment BD intersect outside the circle at *B*. By drawing AC and AD, it can be proven that Δ *ADB* ∼ Δ *CAB*. Therefore,

**Figure 5 **A tangent segment and a secant segment intersecting outside a circle.

This is stated as a theorem.

*Theorem 85:* If a tangent segment and a secant segment intersect outside a circle, then the square of the measure of the tangent segment equals the product of the measures of the secant segment and its external portion.

Also,

*Theorem 86:* If two tangent segments intersect outside a circle, then the tangent segments have equal measures.

**Example 3:** Find *x* in the following figures in 6.

**Figure 6 **A tangent segment and a secant segment (or another tangent segment) intersecting outside a circle.