Let the vertex of an angle ABC be located outside a circle -Maths 9th

Let the vertex of an angle ABC be located outside a circle -Maths 9th
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Given: AD = CE  To prove:  ∠ABC = 1/2(∠DOE – ∠AOC) In △AOD and  △COE AD = CE   (Given) AO = OC and DO = OE (Radii of  same circle) ∴ △AOD ≅ △COE  (By SSS congruence criterion) ⇒  ∠1 = ∠3, ∠2 = ∠4    (CPCT)   ....(i) But OA = OD and OC = OE   ⇒   ∠1 =  ∠2 and  ∠3 =  ∠4  .....(ii) From (i) and (ii), we have   ∠1 =  ∠2 = ∠3 = ∠4  (= x say) Also, OA = OC and OD = DE ⇒ ∠7 = ∠8 (= z say) and ∠5 = ∠6  (= y say) Now, ADEC is a cyclic quadrilateral   ⇒ ∠DAC + ∠DEC = 180° ⇒ x + z + x + y = 180°   ⇒ y = 180° - 2x - z    ...(iii) In △DOE,∠DOE = 180° - 2y  and in △AOC, ∠AOC = 180° - 2z  ∴  ∠DOE - ∠AOC = (180° - 2y) - (180° - 2z) = 2z - 2y = 2z - 2(180° - 2x - z)   (Using (iii)) = 4z + 4x - 360°    ....(iv) Again, ∠BAC + ∠CAD = 180°    ⇒ ∠BAC = 180°– (z + x)  ....(v) Similarly, ∠BCA = 180°– (z + x)   ....(vi) In △ABC,  ∠ABC = 180° - ∠BAC - ∠BCA =  180° - 2[180° - (z - x)]    (Using (v) and (vi)) = 2z + 2x - 180° = 1/2(4z + 4x - 360°)    ....(vii) From (iv) and (vii), we have ∠BAC = 1/2(∠DOE - ∠AOC)
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