# 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:**

Let's construct a diagram based on the given question as shown below:

To prove: ∠ABC = 1/2 (∠DOE - ∠AOC)

Consider ΔAOD and ΔCOE,

OA = OC (Radii of the circle)

OD = OE (Radii of the circle)

AD = CE (Given)

Thus, ∆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) [From equation (4)]

= 4a + 4x - 360° ... (5)

∠BAC + ∠CAD = 180° (Linear pair)

∴ ∠BAC = 180º - ∠CAD = 180° - (a + x) ... (6)

Similarly, ∠ACB = 180° - (a + x) ... (7)

In ΔABC,

∠ABC + ∠BAC + ∠ACB = 180° (Angle sum property of a triangle)

∠ABC = 180° - ∠BAC - ∠ACB

= 180° - (180° - a - x) - (180° - a - x) [From (6) and (7)]

= 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

**☛ Check: **NCERT Solutions Class 9 Maths Chapter 10

**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.

**☛ Related Questions:**

- Prove that the circle drawn with any side of a rhombus as diameter, passes through the point of intersection of its diagonals.
- ABCD is a parallelogram. The circle through A, B and C intersect CD (produced if necessary) at E. Prove that AE = AD.
- AC and BD are chords of a circle which bisect each other. Prove that (i) AC and BD are diameters, (ii) ABCD is a rectangle.
- Bisectors of angles A, B and C of a triangle ABC intersect its circumcircle at D, E and F respectively. Prove that the angles of the triangle DEF are 90° - 1/2 A, 90° - 1/2 B , 90° - 1/2 C.

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