Prove that the quadrilateral formed by the internal angle bisectors of any quadrilateral is cyclic... plz help.
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refer to the attachment
mark the point where bisectors at A and B meet as E and where bisectors at C and D meet as F.
BAE = A/2
ABE = B/2
AEB = 180 - (A/2 + B/2)
like this, CFD = 180 - (C/2 + D/2)
the sum of opposite angles of the quadrilateral formed
= 180 - (A/2 + B/2) + 180 - (C/2 + D/2)
= 360 - (A + B + C + D)/2
= 360 - 180
= 180
the sum of the opposite angles is 180 so it is a cyclic quadrilateral
mark the point where bisectors at A and B meet as E and where bisectors at C and D meet as F.
BAE = A/2
ABE = B/2
AEB = 180 - (A/2 + B/2)
like this, CFD = 180 - (C/2 + D/2)
the sum of opposite angles of the quadrilateral formed
= 180 - (A/2 + B/2) + 180 - (C/2 + D/2)
= 360 - (A + B + C + D)/2
= 360 - 180
= 180
the sum of the opposite angles is 180 so it is a cyclic quadrilateral
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