الأسبوع 8 · تطبيقات التشابه في الدائرة: الوتران المتقاطعان والقاطعان والمماس وعكس ذلك · 3 دروس
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.
Word
Meaning
chord
a segment joining two points of the circle
secant
a line through two points of the circle
tangent
a line that touches the circle at exactly one point
concyclic
points that lie on one circle
Worked examples
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.
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.
5 × 4 = 20 = 2 × ED, so ED = 10.
2
The chord CD has its midpoint at E, and EA = 4, EB = 9 for the chord AB. Find CD.
EC = ED, so EC² = 4 × 9 = 36, EC = 6 and CD = 12.
3
Secants from E outside: EA = 3, AB = 5 and EC = 4. Find ED and CD.
EB = 8. 3 × 8 = 24 = 4 × ED, so ED = 6 and CD = 6 − 4 = 2.
4
Secants from E outside: EA = x, AB = 3, EC = 4 and ED = 10. Find x.