Question

(ii) Prove, if the diagonals of a cyclic quadrilateral are perpendicular to each other, show
that the line passing through the point of intersection of diagonals and midpoint of
a side is perpendicular to the opposite side.​

Answer 1

Let consider midpoint of PQ is T.

Now in a circle line joining the midpoint of chord and center is perpendicular to the chord

so angle PTO is 90° degree

In triangles PTO and QTO,

1), PT = OT (T s the mid point of PQ),

2)TO is common.

3) angle PTO = angle QTO = 90°,

there fore triangles are SAS congruent,

so angle TPO and angle TQO are congruent by C.P.C.T.

similarly angle POT and QOT

so 

angle POT + angle QOT = 90.....given that diagonals are perpendicular

angle POT = angle QOT = 45°

in triangle PTO

angle POT = 45°

angle PTO = 90°

so

angle TPO = 45°

hence 

angle TPO = angle TQO = 45°

now

angle OSU= angle TPO = 45°... (angles subtended by same arc)

angle TQO = angle ORU = 45°... (angles subtended by same arc)

angle TOQ = angle SOU = 45°... (vertically opp. angles),

angle POT = angle ROU = 45°... (vertically opp. angles )

now in triangle OSU

angle OSU= 45°

angle SOU = 45°

hence

angle OUS = (180° - 90°) = 90°...(sum of all angles in triangle)

Hence proved

Answer 2

Proved below.

Step-by-step explanation:

Given:

Let us assume that H is the midpoint of AB . Now as we know that  in a circle line joining the midpoint of chord and center is perpendicular to the chord.  

∠ AHO = 90°            (1)

In Δ AHO and Δ BHO,

AH = BH                  (H s the mid point of AB)

HO = HO                 (common)

∠AHO = ∠BHO       ( 90°)       [from 1]

ΔAHO ≅ ΔBHO      (SAS congruent)

so ∠ AHO ≅ ∠BHO ( by C.P.C.T.)

Similarly, ∠AOH and ∠ BOH, so ∠AOH + ∠BOH = 90°    (given that diagonals are perpendicular)

∠AOH = ∠BOH = 45°

In ΔAHO

∠AOH = 45°, ∠AHO = 90°

so ∠HAO = 45° hence  ∠ HAO = ∠ HBO = 45°  

Now,

∠OSU= ∠HAO = 45°     (angles subtended by same arc)      [2]

∠HBO = ∠ORU = 45°     (angles subtended by same arc)

∠HOB = ∠SOU = 45°     (vertically opp. angles)                      [3]

∠AOH = ∠ROU = 45°     (vertically opp. angles )

Now in Δ OSU,

∠OUS + ∠SOU + ∠OSU= 180°           (angle sum property)

∠OUS + 45° +45° = 180°                     (from Eq 2 and 3)

∠OUS = 180° - 90°

∠OUS = 90°  

Hence proved.