☰ Year plan · Mathematics

Chords, secants and tangents of a circle

Week 8 · Applications of similarity in the circle: intersecting chords, secants and a tangent, and their converses · 3 lessons
1

Two chords, or their extensions, meeting at a point: EA × EB = EC × ED, inside or outside the circle

Mathematics Sec 1 Term 1, Unit 2, Lesson 4
GoalUse the product of the segments of two chords (or secants) that meet at a point.
New words
chord · a segment joining two points on a circlesecant · a line through two points of a circleinscribed angles on one chord · angles on the same side of a chord that look at the same arc are equal
Explanation

Refresher on the circle: a chord joins two points of the circle, a secant is a line through two points of the circle, a tangent touches it in one point, and points on one circle are concyclic. We use one fact: angles on the same side of a chord that look at the same arc are equal. Theorem: if two chords AB and CD (or their extensions) meet at E, inside or outside the circle, then EA × EB = EC × ED. Why: draw AD and CB. The angles ∠EAD and ∠ECB stand on the same chord BD (in the inside case; for E outside one of them is the exterior angle of the cyclic quadrilateral), so they are equal, and the angle at E is common or vertical. So △EAD ~ △ECB, which gives EA/EC = ED/EB, that is EA × EB = EC × ED. Always measure from E to each end: for a point E outside, EB is the whole secant, not AB.

WordMeaning
chorda segment joining two points of the circle
secanta line through two points of the circle
tangenta line that touches the circle at exactly one point
concyclicpoints that lie on one circle
Worked examples
463xABCDE
Two chords AB and CD cross at E inside the circle. EA = 4, EB = 6 and EC = 3. Find ED. EA × EB = 4 × 6 = 24 = EC × ED = 3 × ED, so ED = 8.
Chords AB and CD meet at E inside with EA : EB = 2 : 3, EC = 6 and ED = 5. Let EA = 2k and EB = 3k. Then 2k × 3k = 6 × 5, so 6k² = 30, k² = 5 and k = √5. So EB = 3√5 cm.
x733ABCDE
The lines AB and CD meet at E outside the circle: EC = 3, ED = 6 and AB = 7. Let EA = x, so EB = x + 7. Then x(x + 7) = 3 × 6 = 18, x² + 7x − 18 = 0, (x − 2)(x + 9) = 0. Lengths are positive, so x = 2 and EB = 9.
Class activity

Cord and rope. Draw a circle with a compass, draw two chords that cross, and measure the four pieces with a ruler. Multiply the two pieces of each chord. Are the products equal?

Practice
1
Chords AB and CD cross at E inside a circle: EA = 5, EB = 4 and EC = 2. Find ED.
2
The chord CD has its midpoint at E, and EA = 4, EB = 9 for the chord AB. Find CD.
3
Secants from E outside: EA = 3, AB = 5 and EC = 4. Find ED and CD.
4
Secants from E outside: EA = x, AB = 3, EC = 4 and ED = 10. Find x.
5
A student writes AB × CD = EA × EC for two secants from E. What is wrong?