Question

Let 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. Prove that angle ABC is equal to half the difference of the angles subtended by the chords AC and DE at the centre.

Answer 1

in triangle AOD and COE,
OA=OC
OD=OE
AD=CE
hence,they r congr.t
now,
OAD=OCE (CPCT)
ODA=OEC ( " )
also,
OAD=ODA
i.e. OAD=OCEO=DA=OEC
let them be 'x'
now,
in tri.OAC
OA=OC,so,it's an iso.tri.,let their equal ang's be 'a'
llly
let the equal ang's of ODB be 'y'
 ADEC is a cyclic quad.
now,
AOC=180-2a
DOE=180-2y
DOE-AOC=180-2y-(180-2a)
= -2y+a / 2a-2y
2a-2y= 2a-(360-4x-2a)
       =2a-360+4x+2a
       4a+4x-360
ang.B=180-(180-x-a)-(180-x-a)
         =180-180+x+a-180+x+a
=2x+2a-180
i.e,measure of ang.B/difference = 2x+2a-180/4x+4a-360 =1/2
HENCE THE PROOF.. :)

Answer 2

In ΔAOD and ΔCOE, OA = OC (Radii) OD = OE (Radii) AD = CE (Given) ΔAOD ≅ ΔCOE (SSS congruence rule) ∠OAD = ∠OCE (CPCT) → (1) ∠ODA = Ã∠OEC (CPCT) → (2) and ∠OAD = ∠ODA (OA = OD) → (3) From equations (1), (2), and (3), we get ∠OAD = ∠OCE = ∠ODA = ∠OEC Let∠OAD = ∠OCE = ∠ODA = ∠OEC = x In ΔOAC, OA = OC ⇒ ∠OCA = ∠OAC = a (Angles opposite to equal sides are equal) Consider, ΔODE, OD = OE ∠OED = ∠ODE = y From the figure, ADEC is a cyclic quadrilateral. ∠CAD + ∠DEC = 180° (Sum of opposite angles are supplementary) x + a + x + y = 180° 2x + a + y = 180° y = 180 – 2x – a → (4) Proceed the same way further to get ∠ABC = (1/2)[∠DOE - ∠AOC]