Get Answers to all your Questions

In Fig. 10.19, AB and CD are two chords of a circle intersecting each other at point E. Prove that $\angle$ AEC = $\frac{1}{2}$ (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:

$\angle$ AEC =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,

$\angle$ 1 = 2 $\angle$ 6 … (i)

For arc XA,

$\angle$ 3 = 2 $\angle$ 7 … (ii)

Consider $\triangle$ OAC

OC = OA (Radius of the same circle)

$\angle$ OCA = $\angle$ 4 … (iii) (angles opposite to equal sides in a triangle are equal)

$\angle$ AOC + $\angle$ OCA + $\angle$ 4 = 180° (angle sum property of a triangle)

$\angle$ AOC + $\angle$ 4 + $\angle$ 4 = 180°

$\angle$ AOC = 180° – 2 $\angle$ 4 …(iv)

Consider $\triangle$ DCO

OD = OC (Radius of the same circle)

$\angle$ OCD = $\angle$ 6 … (v) (angles opposite to equal sides in a triangle are equal)

Consider $\triangle$ AEC

$\angle$ AEC + $\angle$ ECA + $\angle$ CAE = 180° (angle sum property of a triangle)

$\angle$ AEC = 180° – ( $\angle$ ECA + $\angle$ CAE)

$\angle$ AEC = 180° – [( $\angle$ ECO + $\angle$ OCA) + ( $\angle$ CAO + OAE)]

$\angle$ AEC = 180° – ( $\angle$ 6 + $\angle$ 4 + $\angle$ 4 + $\angle$ 5) [from (iii) and (v)]

$\angle$ AEC = 180° – (2 $\angle$ 4 + $\angle$ 5 + $\angle$ 6)

$\angle$ AEC = 180° – 2 $\angle$ 4 - $\angle$ 5 - $\angle$ 6

From (iv), $\angle$ AOC = 180° – 2 $\angle$ 4

$\angle$ AEC = $\angle$ AOC - $\angle$ 5 - $\angle$ 6

Also, $\angle$ AOC = $\angle$ 1 + $\angle$ 2 + $\angle$ 3

So, $\angle$ AEC = $\angle$ 1 + $\angle$ 2 + $\angle$ 3 - $\angle$ 5 - $\angle$ 6

In $\triangle$ AOB,

OA = OB (Radius of the same circle)

$\angle$ 5 = $\angle$ 7 (angles opposite to equal sides in a triangle are equal)

$\angle$ AEC = $\angle$ 1 + $\angle$ 2 + $\angle$ 3 - $\angle$ 7 - $\angle$ 6

From (ii), $\angle$ 3 = 2 $\angle$ 7

$\angle$ AEC = $\angle$ 1 + $\angle$ 2 + $\angle$ 3 - 0.5<3 - $\angle$ 6

From (i), $\angle$ 1 = 2 $\angle$ 6

$\angle$ AEC = $\angle$ 1 + $\angle$ 2 + $\angle$ 3 - 0.5 $\angle$ 3 - 0.5 $\angle$ 1

$\angle$ AEC = 0.5 $\angle$ 1 + 0.5 $\angle$ 3 + $\angle$ 2

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

$\angle$ 2 = $\angle$ 8 (vertically opposite angles)

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

$\angle$ AEC = 0.5 ( $\angle$ 1 + $\angle$ 2 + $\angle$ 3 + $\angle$ 8)

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

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

Exams

Colleges

Predictors

Resources

Quick links

Quick Links

Exams

Colleges

Resources

Colleges

Resources

Exams

Exams

Colleges

Resources

Diploma Colleges

Other Exams

Exams

Resources

Upcoming Events

Exams

Ranking

Products & Resources

NCERT Solutions

Top Countries

Student Visas

Exams

Colleges

Exams

Colleges

Upcoming Events

Resources

Exams

Colleges & Courses

Predictors

Resources

Online Courses

Products

Engineering Preparation

Medical Preparation

Top Streams

Specializations

Resources

Top Providers

Exams

Colleges

Predictors

Resources

Resources

Exams

Colleges

Predictors & E-Books

Exams

Animation Courses

Colleges

Resources

Exams

Colleges

Resources

Top Courses & Careers

Colleges

Exams

Resources

Get Answers to all your Questions

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

Answers (1)

Posted by

infoexpert24

Similar Questions

Latest Question

Ask your Query

Welcome Back :)

Register to post Answer

Welcome Back :)

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