• 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 $\angle E$  and $\angle F$  meet at O.

To prove $\angle EOF=90°$

Construction :- Extend FO to meet AB at N .

$$\begin{gathered} Now in ∆FDM and ∆FNB we have \\ \angle DFO = \angle BFN \left(given\right) \\ \angle FDM = \angle FBN (exterior angle of a cyclic quadrilateral) \\ \therefore \angle FMD=\angle FNB (remaining angles of ∆FDM and ∆FNB) \\ But \angle FMD = \angle EMO (vertically opposite angles) \\ \Rightarrow \angle EMO = \angle FNB =\angle ONE..........\left(1\right) \\ Now in ∆EMO and ∆ENO we have \\ \angle OEM = \angle OEN \left(given\right) \\ \angle OME=\angle ONE (from (1\left)\right) \\ \therefore \angle EOM =\angle EON (remaining angles of ∆EMO and ∆ENO) \\ But \angle EOM +\angle EON = 180° (\because MON is a straight line) \\ \Rightarrow \angle EOM +\angle EOM =180° \\ \Rightarrow 2\angle EOM =180° \\ \Rightarrow \angle EOM =90° \\ \therefore \angle EOF =90° \\ Hence internal angle bisectors at E and F are perpendicular \end{gathered}$$