The common tangents PQ and RS to two circles with centre O and O' intersect at M. Prove that ∠OMP + ∠PMR + ∠RMO' = 180°.
Answers
Given:- I)Two circles with centre O and O' respectively.
II) PQ and RS are two common tangents to both the circles
intersecting at M.
To Prove:- ∠OMP + ∠PMR + ∠RMO' = 180°
Construction:- Join OP, OS , O'R and O'Q
Geometrical Property Used:- I)Two tangents drawn from a point to a circle
are equal in length.
II) Side-Side-Side(SSS) Congruency Rule- If all the three sides of
two triangles are equal, then both triangles are congruent.
III) Linear Axiom-The measure of a straight line is 180°, so a linear pair of angles must add up to 180°.
Proof:-
Consider ΔPOM and ΔSOM
PO=SO [ Both are radius of circle having centre O]
OM=OM [ Common side ]
PM=SM [By Geometrical property (I), PM and SM are two tangents from
a point M]
By SSS Congruency,
ΔPOM ≅ ΔSOM
By CPCT( corresponding parts of congruent triangles)
∠PMO = ∠SMO
⇒OM bisects ∠PMS. ..............eqn(1)
Similarly we can prove ΔRO'M ≅ ΔQO'M [ By SSS Congruency ]
By CPCT , ∠RMO' = ∠QMO'.
⇒ O'M bisects ∠RMQ..................................eqn(2)
∠PMS = ∠RMQ [ Vertically Opposite Angles ] ..................eqn(3)
From eqns (1), (2) and (3)
we can say that OM coincides with O'M
⇒OMO' is a straight line.
⇒By Linear Axiom,
∠OMP + ∠PMR + ∠RMO' = 180°
Hence, Proved.
