ABCD is a cyclic quadrilateral whose diagonals AC and BD intersect at right angle at M. Prove that any line passing through M and bisecting any side of the quadrilateral is perpendicular to the opposite side.
 

Let ABCD be a cyclic quadrilateral whose diagonals AC and BD intersect at M at right-angle. MP, bisecting side CD when produced, meets AB in L.
To Prove: ∠PLB = 90°
Proof: ∠AMB = ∠BMC   (given) ∠CMD = ∠DMA = 90°  
DP = CP
∴ PM = PC = PD  [^..^.the median bisecting the hypotenuse of a right-triangle is half of the hypotenuse.]
Now in ΔCPM, MP = CP ⇒ ∠1 = ∠2
But ∠1 = ∠3 [Angles in the same segment of a circle]
and ∠2 = ∠4     (Vertically opposite angles)
.^..  ∠3 = ∠4
But ∠4 + ∠5 = 90° (given)
⇒ ∠3 + ∠5 = 90°
Now in ΔMBL,  ∠MBL + ∠BML + ∠MLB = 180°
⇒ ∠3 + ∠5 + ∠MLB = 180°
or 90° + ∠MLB = 180°
Hence, ∠MLB = 90° ⇒∠PLB = 90°