Question

(ii) Prove, if the diagonals of a cyclic quadrilateral are perpendicular to that the line passing through the point of intersection of diagonals and a side is perpendicular to the opposite side.
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Answer 1

Answer:

Let T be the given midpoint of PQ,

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

so angle PTO is 90 degree

if we take triangles PTO and QYO, we can prove them congruent by SAS

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 = 90...sum of all angles in triangle

hence proved

Step-by-step explanation: