Q. 74.3( 6 Votes )

From a point T ou

Answer :


PT


In ΔOPT and ΔOQT
OPT = OQT = 90°


OP = OQ (radius)


OT = OT (common side)


ΔOPT and ΔOQT are congruent by RHS.


So, PTO = QTO (by CPCT) ….(1)


Also, PT = QT (by CPCT) …(2)


Now,
In ΔPCT and Δ QCT,
PT = QT (From eq (2))


PTC = QTC (From eq(1))


CT = CT (common side)


ΔPCT and Δ QCT are congruent by SAS property.


So, PCT = QCT (by CPCT) …..(3)


As PQ is a line segment,
PCT + QCT = 180°


3 + 4 = 180°


3 + 3 = 180°


23 = 180°


3 = 90°


Therefore, 3 = 4 = 90°


Hence proved that OT is the right bisector of PQ.


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