PA and PB are tangents from P to the circle with centre O. At the poin...
Here, PA=PB , KA = KM , NB = NM,[length of tangents from external points are equal]therefore, KA + NB =KM +MNor KA + BN =KM + MN = KN
PA and PB are tangents from P to the circle with centre O. At the poin...
Proof:
Let's prove that KN = AK = BN.
1. Tangents are perpendicular to radii:
Since PA and PB are tangents to the circle with center O, they are perpendicular to the radii drawn from the center O to the points A and B.
2. Triangles AOM and BOM:
The triangles AOM and BOM are right-angled triangles because OA and OB are radii of the circle, and AM and BM are tangents to the circle. Therefore, angles OAM and OBM are right angles.
3. Triangles AOM and BOM are congruent:
Since angles OAM and OBM are right angles and OA = OB (both are radii of the circle), we can conclude that the triangles AOM and BOM are congruent by the hypotenuse-leg congruence theorem.
4. Angle MOB = angle MOA:
Since the triangles AOM and BOM are congruent, their corresponding angles are also congruent. Therefore, angle MOB = angle MOA.
5. Triangles AKN and BKN:
The triangles AKN and BKN are right-angled triangles because AK and BN are tangents to the circle, and KN is a common tangent. Therefore, angles AKM and BNM are right angles.
6. Triangles AKN and BKN are congruent:
Since angles AKM and BNM are right angles and KN = KN (common side), we can conclude that the triangles AKN and BKN are congruent by the hypotenuse-leg congruence theorem.
7. AK = KN = BN:
Since the triangles AKN and BKN are congruent, their corresponding sides are also congruent. Therefore, AK = KN = BN.
Thus, it is proven that KN = AK = BN.
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