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 ∠ABC is equal to half the difference of the angles subtended by the chords AC and DE at the centre.

Answer 1

Data : The vertex of angle ABC be located outside a circle and let the sides of the angle intersect equal chords AD and CE with the circle. 

To Prove: ∠ABC is equal to half the difference of the angles subtended by the chords AC and DE at the centre. OR 

∠ABC= \(\frac{1}{2}\)[∠DOE – ∠AOC]. 

Construction: OA, OC, OE, OD are joined. 

Proof: In ∆AOD and ∆COE, 

OA = OC, 

OD = OE radii of same circle. 

AD = CE (Data) 

∴ ∆AOD ≅ ∆COE (SSS Postulate) 

∠OAD = ∠OCE ……….. (i) 

∴∠ODA = ∠OEC …………. (ii) 

OA = OD 

∴ ∠OAD = ∠ODA …………. (iii) 

From (i) and (ii), 

∠OAD = ∠OCE = ∠ODA 

= ∠OEC = x°. 

In ∆ODE, 

OD = OE 

∠ODE = ∠OED = y°. 

ADEC is a cyclic quadrilateral. 

∴ ∠CAD + ∠DEC = 180° 

x + a + x + y = 180 

2x + a + y = 180 

y = 180 – 2x – a ……….. (iv) 

But, ∠DOE = 180 – 2y