Question

# Question 8 In the figure, AB and CD are two chords of a circle intersecting each other at point E. Prove that ∠AEC=12 (angle subtended by arc CXA at centre + angle subtended by arc DYB at the centre).

Solution

## Construction: Extend the line DO and BO at the points l and H on the circle. Also, join AC. We know that, in a circle, the angle subtended by an arc at the centre is twice the angle subtended by it in the remaining part of the circle. ∴∠1=2∠6 ...(i) and ∠3=2∠7 ...(ii) In ΔAOC,OC=OA [both are the radius of circle] ∠OCA=∠4 [angles opposite to equal sides are equal] Also, ∠AOC+∠OCA+∠4=180∘ [by angle sum property of triangle] ⇒∠AOC+∠4+∠4=180∘ ⇒∠AOC=180∘−2∠4 ...(iii) Now, in ΔAEC,∠AEC+∠ECA+∠CAE=180∘ [by angle sum propertyof a triangle] ⇒∠AEC=180∘−(∠ECA+∠CAE) ⇒∠AEC=180∘−[(∠ECO+∠OCA)+∠CAO+∠OAE] =180∘−(∠6+∠4+∠4+∠5) [in ΔOCD,∠6=∠ECO, because, angles opposite to equal sides are equal] =180∘−(2∠4+∠5+∠6) =180∘−(180∘−∠AOC+∠7+∠6) [From Eq. (iii) and in ΔAOB,∠5=∠7, as angles opposite to equal sides are equal] =∠AOC−∠32−∠12                [From Eqs. (i) and (ii)] =∠AOC−∠12−∠22−∠32+∠22 [adding and subtracting ∠22] =∠AOC−12(∠1+∠2+∠3)+∠82 [∵∠2=∠8 (vertically opposite angles)] =∠AOC−∠AOC2+∠DOB2 ⇒∠AEC=12(∠AOC+∠DOB) =12 [Angle sutended by arc CXA at the centre + angle subtended by arc DYB at the centre.]MathematicsNCERT ExemplarStandard IX

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