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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=26 ...(i)
and 3=27 ...(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=18024 ...(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(24+5+6)
=180(180AOC+7+6)
[From Eq. (iii) and in ΔAOB,5=7, as angles opposite to equal sides are equal]
=AOC3212                [From Eqs. (i) and (ii)]
=AOC122232+22 [adding and subtracting 22]
=AOC12(1+2+3)+82 [2=8 (vertically opposite angles)]
=AOCAOC2+DOB2
AEC=12(AOC+DOB)
=12
[Angle sutended by arc CXA at the centre + angle subtended by arc DYB at the centre.]

Mathematics
NCERT Exemplar
Standard IX

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