Ex 10.6, 4 (Optional) - Let the vertex of angle ABC be located outside

Ex 10.6, 4 (Optional) - Chapter 10 Class 9 Circles - Part 2
Ex 10.6, 4 (Optional) - Chapter 10 Class 9 Circles - Part 3 Ex 10.6, 4 (Optional) - Chapter 10 Class 9 Circles - Part 4 Ex 10.6, 4 (Optional) - Chapter 10 Class 9 Circles - Part 5 Ex 10.6, 4 (Optional) - Chapter 10 Class 9 Circles - Part 6

Ex 10.6, 4 (Optional) - Chapter 10 Class 9 Circles - Part 7 Ex 10.6, 4 (Optional) - Chapter 10 Class 9 Circles - Part 8 Ex 10.6, 4 (Optional) - Chapter 10 Class 9 Circles - Part 9 Ex 10.6, 4 (Optional) - Chapter 10 Class 9 Circles - Part 10

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Question 4 – Method 1 Let the vertex of an angle ABC be located outside a circle and let the sides of the angle intersect equal chords AD and CE with the circle. Prove that ∠ABC is equal to half the difference of the angles subtended by the chords AC and DE at the center. Given: Equal chords AD and CE To Prove: ∠ABC = 1/2 (∠AOC − ∠DOE) Construction: Join AC and DE Proof: Given, AD = CE ∴ ∠AOD = ∠COE (Equal chords subtend equal angle at the center) Let ∠AOD = ∠COE = 𝑥 Now, around point O, Sum of angles around a point = 360° ∠1 + ∠2 + ∠AOD + ∠COE = 360° (∠1 + ∠2) + 𝑥 + 𝑥 = 360° 2𝑥 = 360° − (∠1 + ∠2) 𝑥 = (360°)/2 − ((∠1 + ∠2))/2 𝑥 = 180° − 1/2 (∠1 + ∠2) Now, In ∆COE OC = OE ∠OEC = ∠OCE In ∆COE, ∠COE + ∠OCE + OEC = 180° x + ∠OCE + ∠OCE = 180° x + 2 ∠OCE = 180° 2 ∠OCE = 180° − 𝑥 ∠OCE = (180°)/2 − 𝑥/2 ∠ OCE = 90° − 𝑥/2 (Both are radius) (Angles Opposite to equal sides of a triangle are equal ) (Angle sum property) (As ∠OEC = ∠OCE) Similarly, ∠ OAD = 90° − 𝑥/2 Now, For line BE ∠ BCO + ∠ OCE = 180° ∠BCO = 180° − ∠ OCE ∠BCO = 180° − (90° – 𝑥/2) ∠BCO = 90° + 𝑥/2 Similarly, ∠BAO = 90° + 𝑥/2 (Linear Pair) Now In quadrilateral ABCO, ∠B + ∠BCO + ∠ BAO + ∠1 = 360° ∠ B + 90° + 𝑥/2 + 90° + 𝑥/2 + ∠1 = 360° ∠ B + 180° + x + ∠1 = 360° ∠ B + 180° + 180° − 𝟏/𝟐 (∠1 + ∠2) + ∠ 1 = 360° ∠ B + 360° − 1/2 ∠1 − 1/2 ∠2 + ∠1 = 360° ∠ B + 1/2 ∠1 − 1/2 ∠2 = 360° − 360° ∠ B + 1/2 ∠1 − 1/2 ∠2 = 0 (Sum of angles of a quadrilateral is 360° ) From (1) - Putting 𝑥 = 180°− 1/2 (∠1 + ∠2) ∠ B = 1/2 ∠2 − 1/2 ∠1 ∠B = 1/2 (∠2 − ∠1) Hence Proved Question 4 – Method 2 Let the vertex of an angle ABC be located outside a circle and let the sides of the angle intersect equal chords AD and CE with the circle. Prove that ∠ABC is equal to half the difference of the angles subtended by the chords AC and DE at the center. Given: Equal chords AD and CE To Prove: ∠ABC = 1/2 (∠AOC − ∠DOE) Construction: Join AC, DE and AE, DC Proof: We know that, Angle subtended by arc at the center is double angle subtended by it at any other point For arc DE Angle subtended at the center = ∠ DOE = ∠ 2 …(1) Angle subtended on circle = ∠ DCE ∴ ∠ DCE = (∠ 2)/2 For arc AC Angle subtended at the center = ∠ AOC = ∠ 1 Angle subtended on circle = ∠ AEC ∴ ∠ AEC = (∠ 1)/2 …(2) In Δ ABE Exterior angle is equal to sum of interior opposite angles ∴ ∠ DAE = ∠ ABC + ∠ AEC ∴ ∠ DAE = ∠ ABC + (∠1)/2 Also, ∠ DCE = ∠ DAE So, our equation becomes ∠ DCE = ∠ ABC + (∠1)/2 (∠2)/2 = ∠ ABC + (∠1)/2 (∠2)/2 – (∠1)/2 = ∠ ABC (Angles subtended by a chord on same side of triangle are equal) (From (1): ∠ DCE = (∠ 2)/2) 1/2 (∠2 – ∠1) = ∠ ABC 1/2 (∠ AOC – ∠ DOE) = ∠ ABC Hence proved

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.