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.

Question

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 &#x2220; ABC is equal to half the difference of the angles subtended by the chords AC and DE at the centre.

Philoid · Accepted Answer

In ΔAOD and ΔCOE,

OA = OC (radii)

OD = OE (Radii)

AD = CE (Given)

Therefore,

By SSS congruence,

ΔAOD ΔCOE

∠OAD = ∠ OCE (By CPCT) (i)

∠ODA = ∠OEC (By CPCT) (ii)

∠OAD = ∠ODA (iii)

Using (i), (ii) and (iii), we get

∠OAD = ∠OCE = ∠ODA = ∠OEC = x

Now,

In triangle ODE,

OD = OE

Hence, ABCD is a quadrilateral

∠CAD + ∠DEC = 180^o

a + x + x + y = 180^o

2x + a + y = 180^o (iv)

∠DOE = 180^o – 2y

And,

∠COA = 180^o - 2a

∠DOE - ∠COA = 2a – 2y

= 2a – 2 (180^o - 2x – a)

= 4a + 4x – 360^o (v)

∠BAC = 180^o – (a + x)

And

∠ACB = 180 – (a - x)

In ΔABC,

∠ABC + ∠BAC + ∠ACB = 180^o (Angle sum property of triangle)

∠ABC = 2a + 2x – 180^o

∠ABC = 1/2 (4a + ax – 360⁰) [using (v)]

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.