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  • In Fig. 10.19, AB and CD are two chords of a circle intersecting each other at point E. Prove that angle AEC = 1 divided 2 (Angle subtended by arc CXA at centre + angle subtended by arc DYB at the centre).

In Fig. 10.19, AB and CD are two chords of a circle intersecting each other at point E. Prove that \angleAEC =\frac{1}{2} (Angle subtended by arc CXA at centre + angle subtended by arc DYB at the centre).

Answers (1)

Given: AB and CD are two chords of a circle intersecting each other at point E

To prove:

\angleAEC =0.5 (Angle subtended by arc CXA at centre + angle subtended by arc DYB at the centre)

\angle AEC=0.5\left ( \angle AOC+\angle DOB \right )

Proof:

Extend the lines DO to L and BO to point X on the circle. Join AC.

Now we know that angle subtended at the centre by an arc is twice the angle subtended by it at any part of the circle.

For arc LC,

\angle1 = 2\angle6                    … (i) 

For arc XA,

\angle3 = 2\angle7                    … (ii)

Consider \triangleOAC

OC = OA   (Radius of the same circle)

\angleOCA = \angle4                … (iii)  (angles opposite to equal sides in a triangle are equal)

\angleAOC + \angleOCA + \angle4 = 180° (angle sum property of a triangle)

\angleAOC + \angle4 + \angle4 = 180°

\angleAOC = 180° – 2 \angle4 …(iv)

Consider \triangleDCO

OD = OC    (Radius of the same circle)

\angleOCD = \angle6                … (v)   (angles opposite to equal sides in a triangle are equal)

Consider \triangleAEC

\angleAEC + \angleECA + \angleCAE = 180°  (angle sum property of a triangle)

\angleAEC = 180° – (\angleECA + \angleCAE)

\angleAEC = 180° – [(\angleECO + \angleOCA) + (\angleCAO + OAE)]

\angleAEC = 180° – (\angle6 + \angle4 + \angle4 + \angle5)          [from (iii) and (v)]

\angleAEC = 180° – (2\angle4 + \angle5 + \angle6)

\angleAEC = 180° – 2\angle4 - \angle5 - \angle6

From (iv), \angleAOC = 180° – 2 \angle4

\angleAEC = \angleAOC - \angle5 - \angle6

Also, \angleAOC = \angle1 + \angle2 + \angle3

So, \angleAEC = \angle1 + \angle2 + \angle3 - \angle5 - \angle6

In \triangleAOB,

OA = OB (Radius of the same circle)

\angle5 =\angle7  (angles opposite to equal sides in a triangle are equal)

\angleAEC = \angle1 +\angle2 + \angle3 - \angle7 - \angle6

From (ii), \angle3 = 2\angle7  

\angleAEC = \angle1 + \angle2 + \angle3 - 0.5<3 - \angle6

From (i), \angle1 = 2\angle6 

\angleAEC = \angle1 + \angle2 + \angle3 - 0.5\angle3 - 0.5\angle1

\angleAEC = 0.5\angle1 + 0.5\angle3 + \angle2

\angle AEC = \frac{1}{2}\angle 1 + \frac{1}{2}\angle 3 + \frac{1}{2}\angle 2 + \frac{1}{2}\angle 2

\angle2 = \angle8                      (vertically opposite angles)

\angle AEC = \frac{1}{2}\angle 1 + \frac{1}{2}\angle 3 + \frac{1}{2}\angle 2 + \frac{1}{2}\angle 8

\angleAEC = 0.5 (\angle1 + \angle2 + \angle3 + \angle8)

\angle AEC=0.5\left ( \angle AOC+\angle DOB \right )

\angleAEC =0.5 (Angle subtended by arc CXA at centre + angle subtended by arc DYB at the centre)

Hence proved

Posted by

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