Q. 43.8( 62 Votes )

Let the vertex of

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 = 180o


a + x + x + y = 180o


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


∠DOE = 180o – 2y


And,


∠COA = 180o - 2a


∠DOE - ∠COA = 2a – 2y


= 2a – 2 (180o - 2x – a)


= 4a + 4x – 360o (v)


∠BAC = 180o – (a + x)


And


∠ACB = 180 – (a - x)


In ΔABC,


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


∠ABC = 2a + 2x – 180o


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

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