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 centre
Solution:
We can solve this question by using the following properties

The angle subtended by an arc at the center is double the angle subtended by it at any point on the remaining part of the

A quadrilateral ACED is called cyclic if all the four vertices of it lie on a

The sum of either pair of opposite angles of a cyclic quadrilateral is 180°.

Using SideSideSide (SSS criteria) and Corresponding parts of congruent triangles (CPCT) we prove the statement.
Consider ΔAOD and ΔCOE,
OA = OC (Radii of the circle)
OD = OE (Radii of the circle)
AD = CE (Given)
∠AOD ≅ ∠COE (SSS Congruence Rule)
∠OAD = ∠OCE (By CPCT) ... (1)
∠ODA = ∠OEC (By CPCT) ... (2)
Also,
∠OAD = ∠ODA (As OA = OD) ... (3)
From Equations (1), (2), and (3), we obtain
∠OAD = ∠OCE = ∠ODA = ∠OEC
Let ∠OAD = ∠OCE = ∠ODA = ∠OEC = x
In ΔOAC,
OA = OC
∴ ∠OCA = ∠OAC (Angle a)
In ΔODE,
OD = OE
∠OED = ∠ODE (Angle y)
ADEC is a cyclic quadrilateral.
∴ ∠CAD + ∠DEC = 180° (Opposite angles are supplementary)
x + a + x + y = 180°
2x + a + y = 180°
y = 180°  2x  a ... (4)
However, ∠DOE = 180°  2y and, ∠AOC = 180° − 2a
∠DOE  ∠AOC = 2a  2y = 2a  2 (180°  2x  a)
= 4a + 4x  360° ... (5)
∠BAC + ∠CAD = 180° (Linear pair)
∴ ∠BAC = 180º  ∠CAD = 180°  (a + x)
Similarly, ∠ACB = 180°  (a + x)
In ΔABC,
∠ABC + ∠BAC + ∠ACB = 180° (Angle sum property of a triangle)
∠ABC = 180°  ∠BAC  ∠ACB
= 180°  (180°  a  x)  (180°  a  x)
= 2a + 2x 180°
= 1/2 [4a + 4x  360°]
Using Equation (5)
∠ABC = 1/2 (∠DOE  ∠AOC )
Hence it is proved that ∠ABC is equal to half the difference of the angles subtended by the chords AC and DE at the center
Video Solution:
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 centre
Maths NCERT Solutions Class 9  Chapter 10 Exercise 10.6 Question 4:
Summary:
The vertex of an angle ABC is located outside a circle, and the sides of the angle intersect equal chords AD and CE with the circle. We have proved that ∠ABC is equal to half the difference of the angles subtended by the chords AC and DE at the center.