Question

in the given figure, O is the circumcentre of triangle ABC. If BD and CD are the bisectors of angle ABC and angle ACB respectively, prove that 4 angle BDC - angle BOC = 360°.

Answer 1

Let Angle BAC be x.
Therefore, Angle BOC = 2x(Angle subtended at the centre)
In ∆ABC,
Angle A + Angle B + Angle C = 180°
x + Angle B + Angle C = 180°
Angle(B +C) =180° - x

So, ½(Angle B + Angle C) = ½(180°-x)
½(Angle B + Angle C) = 90° - ½x
In ∆DBC,
Angle DBC = ½Angle ABC
Angle DCB = ½Angle ACB
Therefore,
½(Angle B + Angle C) + Angle BDC = 180°
90° - ½x + Angle BDC = 180°
Angle BDC = 180°-(90°-½x)
Angle BDC = 90°+½x

Now, according to question,
4 angle BDC - angle BOC = 360°
Substituting values
4(90°+½x) - 2x = 360°
360°+2x-2x = 360°
360°=360°

Hence,proved.