• If the opposite sides of a cyclic quadrilateral are extended to meet, say at points E and F, then the internal angle bisectors of the angles formed at points E and F are perpendicular.

Given ABCD is a cyclic quadrilateral .AB and DC are produced to meet at E whereas AD and BC are produced to meet at F.
Internal angle bisectors of E  and F  meet at O.

To prove EOF=90°

Construction :- Extend FO to meet AB at N .

Now in FDM and FNB we haveDFO = BFN (given)FDM = FBN (exterior angle of a cyclic quadrilateral)FMD=FNB  (remaining angles of FDM and FNB)But FMD = EMO (vertically opposite angles)EMO = FNB =ONE..........(1)Now in EMO and ENO we haveOEM = OEN (given)OME=ONE      (from (1))EOM =EON (remaining angles of EMO and ENO)But EOM +EON = 180° (MON is a straight line)EOM +EOM =180°2EOM =180°EOM =90°EOF =90°  Hence internal angle bisectors at E and F are perpendicular

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