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Circles Tangents Problems (view source)
Revision as of 06:15, 13 August 2014
, 06:15, 13 August 2014→problem 3 [Ex-15.2 B.7]
Circles <math>C_{1}</math> and <math>C_{2}</math> touch internally at a point A and AB is a chord of the circle<math>C_{1}</math> intersecting <math>C_{2}</math> at P, Prove that AP= PB.<br>
Circles <math>C_{1}</math> and <math>C_{2}</math> touch internally at a point A and AB is a chord of the circle<math>C_{1}</math> intersecting <math>C_{2}</math> at P, Prove that AP= PB.<br>
[[Image:problem 3 on circle.png|300px]]
[[Image:problem 3 on circle.png|300px]]
Concepts used
==Concepts used==
1. The radii of a circle are equal
1. The radii of a circle are equal
2.Properties of isosceles triangle.
2.Properties of isosceles triangle.
3.SAS postulate
3.SAS postulate
4.Properties of congruent triangles.
4.Properties of congruent triangles.
Prerequisite knowledge
==Prerequisite knowledge==
1. The radii of a circle are equal.
1. The radii of a circle are equal.
2. In an isosceles triangle angles opposite to equal sides are equal.
2. In an isosceles triangle angles opposite to equal sides are equal.
3.All the elements of congruent triangles are equal.
3.All the elements of congruent triangles are equal.
==Algoritham==
In ∆AOB
AO=BO [Radii of a same circle]
∴ ∠OAB = ∠OBA --------------I [∆AOB is an isosceles ∆}
Then,
In ∆AOP and ∆BOP,
AO = BO [Radii of a same circle]
OP=OP [common side]
∠OAP = ∠OBP [ from I]
∴ ∆AOP ≅ ∆BOP [SAS postulate]
∴ AP = BP [corresponding sides of congruent triangles ]