Here we will prove that if a line touches a circle and from the point of contact a chord is down, the angles between the tangent and the chord are respectively equal to the angles in the corresponding alternate segments.
Given: A circle with centre O. Tangent XY touches the circle at the point M. Through M, a chord MN is drawn. Let MN subtend ∠MSN and ∠MTN in the major and minor segments respectively.
To prove: ∠NMY = ∠MSN and ∠NMX = ∠MTN
Construction: Draw the diameter MOR. Join N to R.
Proof:
Statement: |
Reason |
1. ∠RMY = 90° ⟹ ∠RMN + ∠NMY = 90° ⟹ ∠NMY = 90° - ∠RMN |
1. Diameter ⊥ Tangent. |
2. In ∆RMN, ∠MNR = 90° |
2. Angle in a semicircle is 90°. |
3. ∠NRM + ∠RMN = 90° |
3. In a right-angled triangle, sum of the two acute angles is 90°. |
4. ∠NRM = ∠MSN |
4. Angles in the same segment are equal. |
5. ∠MSN + ∠RMN = 90° ⟹ ∠MSN = 90° - ∠RMN |
5. From statements 3 and 4. |
6. ∠NMY = ∠MSN |
6. From statements 1 and 5. |
7. ∠NMY + ∠NMX = 180° |
7. Linear pair. |
8. ∠MSN + ∠MTN = 180° |
8. Opposite angles of a cyclic quadrilateral are supplementary. |
9. ∠NMY + ∠NMX = ∠MSN + ∠MTN |
9. From 7 and 8. |
10. ∠NMX = ∠MTN. |
10. ∠NMY = ∠MSN from statement 6. |
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