question_answer

Prove that the tangent drawn at the mid-point of an arc of a circle is parallel to the chord joining the end points of the arc.

Answer:

Given, C is the mid-point of the minor arc PQ and O is the centre of the circle and AB is tangent to the circle through point C.

Construction: Draw PC and QC.

To prove:        \[PQ\parallel AB\]

Proof: It is given that C is the mid-point of the arc PQ.

So,      Minor arc PC = Minor arc QC

\[\Rightarrow PC=QC\]

Hence \[\Delta \,PQC\] is an isosceles triangle.

Thus the perpendicular bisector of the side PQ of \[\Delta \,PQC\] passes through vertex C.

But we know that the perpendicular bisector of a chord passes through centre of the circle.

So, the perpendicular bisector of PQ passes through the center O of the circle.

Thus, the perpendicular bisector of PQ passes through the points O and C.

\[\Rightarrow PQ\bot OC\]                                                            ?(i)

AB is perpendicular to the circle through the point C on the circle

\[\Rightarrow AB\bot OC\]                                                            ?(ii)

From equations (i) and (ii), the chord PQ and tangent AB of the circle are perpendicular to the same line OC.

Hence,              \[AB\parallel PQ\]

or                     \[PQ\parallel AB\]                                   Hence Proved.