Here we will prove that from any point outside a circle two tangents can be drawn to it and they are equal in length.
Given: O is the centre of a circle and T is a point outside the circle.
Construction: Join O and T. Draw a circle with TO as diameter which cuts the given circle at M and N. Join T to M and N.
To prove: TM and TN are tangent to the circle and TM = TN.
Proof:
Statement |
Reason |
1. ∠TMO = 90°. |
1. Angle in a semicircle is a right angle. |
2. TM ⊥ OM. |
2. From statement 1. |
3. Therefore, TM is a tangent to the given circle. |
3. Tangent ⊥ radius drawn through point of contact. |
4. Similarly, TN is a tangent to the given circle. |
4. Proceeding as above. |
5. In ∆TOM and ∆TON, (i) OM = ON. (ii) ∠OMT = ∠ONT = 90°. (iii) TO = TO. |
5. (i) Radii of the same circle. (ii) Radius ⊥ tangent. (iii) Common side. |
6. ∆TOM ≅ ∆TON. |
6. By RHS criterion. |
7. TM = TN. |
7. CPCTC. |
Note:
1. The two tangents subtend equal angles at the centre of the circle.
∠TOM = ∠TON, as ∆TOM ≅ ∆TON.
2. The two tangents are equally inclined to the line joining the point to the centre of the circle.
∠MTO = ∠NTO, as ∆TOM ≅ ∆TON.
Alternate Segments
In the given below figure, the chord MN divides the circle into two segments. The tangent XY is drawn that touches the circle N.
The alternate segment for ∠MNY is the segment MAN and that for ∠MNX is the segment MBN.
The angle in the alternate segment for ∠MNY is ∠MAN and that for ∠MNX is ∠MBN.
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