Question -
Answer -
Consider the diagram
In ΔAOD andΔCOE,
OA = OC (Radii of the same circle)
OD = OE (Radii of the same circle)
AD = CE (Given)
∴ ΔAOD ≅ ΔCOE (SSS congruence rule)
∠OAD = ∠OCE (By CPCT) … (1)
∠ODA = ∠OEC (By CPCT) … (2)
Also,
∠OAD = ∠ODA (As OA = OD) … (3)
From equations (1), (2), and (3), we obtain
∠OAD = ∠OCE = ∠ODA = ∠OEC
Let ∠OAD = ∠OCE = ∠ODA = ∠OEC = x
In Δ OAC,
OA = OC
∴ ∠OCA = ∠OAC (Let a)
In Δ ODE,
OD = OE
∠OED = ∠ODE (Let y)
ADEC is a cyclic quadrilateral.
∴ ∠CAD + ∠DEC = 180° (Opposite angles aresupplementary)
x + a + x + y =180°
2x + a + y =180°
y = 180º − 2x − a …(4)
However, ∠DOE = 180º − 2y
And, ∠AOC = 180º − 2a
∠DOE − ∠AOC = 2a − 2y = 2a − 2 (180º − 2x − a)
= 4a +4x −360° … (5)
∠BAC + ∠CAD = 180º (Linear pair)
⇒ ∠BAC = 180º − ∠CAD = 180º− (a + x)
Similarly, ∠ACB = 180º− (a + x)
In ΔABC,
∠ABC + ∠BAC + ∠ACB = 180º(Angle sum property of a triangle)
∠ABC = 180º − ∠BAC − ∠ACB
= 180º − (180º − a − x)− (180º − a −x)
= 2a +2x −180º
= 
[4a + 4x − 360°]
∠ABC = 
[∠DOE − ∠AOC] [Using equation (5)]