Circle Theorems and Secant Relationships

Angle Formed by Intersecting Secants

• When two secants intersect inside a circle, the measure of the angle formed is half the sum of the intercepted arcs.

  • For a central angle, it equals the arcs measure.

  • Since the angle is outside the center, it intersects two arcs.

  • $Measure \ of \ angle \ 1 = \frac{1}{2}(AD + BC)$

  • This si the average of the two arcs.

• Vertical angles share the same property:

  • The vertical angle to angle one also equals half the sum of arcs AD and BC.

  • $Measure \ of \ angle \ CED = \frac{AD + BC}{2}$

• Example:

  • Given arc AD = 155 degrees and arc BC = 61 degrees.

  • Find the measure of angle one.

  • $Measure \ of \ angle \ 1 = \frac{155 + 61}{2} = \frac{216}{2} = 108 \ degrees$

• Finding other angles:

  • If angle one is 108 degrees, its vertical angle is also 108 degrees.

  • Supplementary angles (linear pair) add up to 180 degrees.

  • The supplement of 108 degrees is 72 degrees.

  • The vertical angle to this supplement is also 72 degrees.

Theorem: Measure of an Angle Formed by Two Secant Lines

• The measure of an angle formed by two lines (secants) that intersect outside a circle is half the difference of the measures of 4356the intercepted arcs. *Illustration:

  • Consider angle one formed by two secants outside the circle.

  • $Measure \ of \ angle \ 1 = \frac{Major \ Arc - Minor \ Arc}{2}$

Algebraic Application of the Theorem

• Given:

  • Major arc = $6x - 6$

  • Minor arc = $3x - 5$

  • Measure of angle KJN = 34 degrees.

• Applying the theorem:

  • $34 = \frac{(6x - 6) - (3x - 5)}{2}$

• Simplifying:

  • $34 = \frac{6x - 6 - 3x + 5}{2}$

  • $34 = \frac{3x - 1}{2}$

• Solving for x:

  • Multiply both sides by 2: $68 = 3x - 1$

  • Add 1 to both sides: $69 = 3x$

  • Divide by 3: $x = 23$

    • Correction the previous calculation contained an error

    • $34 = \frac{6x - 6 - 3x + 5}{2}$

    • $34 = \frac{3x - 1}{2}$

    • Multiply both sides by 2: $68 = 3x - 1$

    • Add 1 to both sides: $68 + 1 = 3x$

    • $69 = 3x$

    • Divide by 3: $x = 23$

• Corrected Solution

  • Multiply both sides by 2: $2 * 34 = 3x - 6 + 5$

  • $68 = 3x -1$

  • $68 + 1 = 3x$

  • $69 = 3x$

  • $x = 23$

• Another Correction identified:
  *$34 = \frac{(6x - 6) - (3x + 5)}{2}$
  *$34 = \frac{6x - 6 - 3x - 5}{2}$
  *$34 = \frac{6x - 3x - 6 - 5}{2}$
  *$34 = \frac{3x - 11}{2}$

  • Multiply both sides by 2: $2 * 34 = 3x - 11$

  • $68 = 3x -11$

  • $68 + 11 = 3x$

  • $79 = 3x$

  • $x = 79/3$

  • $x = 26.33$

• Given measure of arc LM = $6x + 6$

• Find measure of LM by substituting x = 19:

  • $Measure \ of \ LM = 6(19) + 6 = 114 + 6 = 120$ degrees

Setting up Equations with Intersecting Secants

• If the angle is 107, set up the equation to find arc WX:

  • $107 = \frac{74 + x}{2}$

Key Rules to Remember

• Angle inside the circle: Add the intercepted arcs and divide by two.

• Angle outside the circle: Subtract the intercepted arcs and divide by two.

• Outside angle is usually smaller.

• Inside angle is generally larger.

Finding Arcs and Angles

• To find arc 'a':

  • Arc a = 82 degrees.

• Measure of angle s:

  • $Measure \ of \ angle \ s = \frac{152 - 82}{2} = \frac{70}{2} = 35 \ degrees$

Similar Triangles Formed by Secants

• When two secant lines cut through a circle, the triangles formed (one at the vertex and the other by the vertical angle) are similar, not congruent.

• Reason

  • Angle C and Angle D are congruent as inscribed angles intercepting the same arc. This angle is half of the measure of the inscribed arc.

  • Triangle similarity is proved through AA (Angle-Angle) theorem.

Segments of Two Secants Rule

• When secants intersect, the product of the segments on one secant is proportional to the product of the segments on the other. *Illustration

  • Segments x and 4 on one secant and segments 6 and 5 on the other.

  • \6 * 5 = 4 * x

  • $30 = 4x$

  • $x = 7.5$

Rules for Segments of Intersecting Secants and Tangents

• Two segments on one secant, their product equals to the product of the segments on the other.

• The whole segment times the outside segment equals the whole segment times the outside segment for two secants.

• If one line is a tangent, then $x^2$ equals the whole segment times the outside segment of the secant. *Formulas to Remember

  • $a * b = c * d$

  • Whole segment times outside segment equals the whole segment times the outside segment.

  • $MN * N = (B + Q) * Q$

Application

• Archaeologists find circular walls and need to find a length inside the circle.

• They want to apply : The whole thing times the part outside.

• The same concepts are applied in solving word problems, involving the use of established mathematical theorems.

Solving for Unknown Segment Lengths

*Formula
* $Tangent^2 = (Whole \ Secant) * (External \ Part \ of \ Secant)$

• Finding the value of a

  • Set up: $6^2 = (a + 9) * a$

  • $36 = a^2 + 9a$

  • $a^2 + 9a - 36 = 0$

• Factor the quadratic:

  • Find two numbers that multiply to -36 and add to 9.

  • These numbers are 12 and -3 (12 * -3 = -36 and 12 + -3 = 9).

  • $(a + 12)(a - 3) = 0$

• Solve for a:

  • a = -12 or a = 3

  • Since distance cannot be negative, a = 3.

Calculating Segment Lengths

• To find segment EC:

  • Let EC be x.

  • Then, 4 * 5 = 2 * x ( 20 = 2x, x= 10)