Question

1) The diagonals of a cyclic square
are perpendicular to each other. So
prove that the line passing through the
midpoint of one side and the intersection
of the ears is perpendicular to the
opposite side.​

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 TOO 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 TO = angle ORU =

subtended by same arc

45. angles

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

Answer 2

Answer:

Using the figure in the image

Step-by-step explanation:

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