Question

prove that the centre of a circle touching two intersecting lines lies on the angle bisetor of the lines

Answer 1

Answer:

A circle wit centre '' O '''.

⇒ And , also O touches Line - l₁ and l₂ .

⇒ l₁ and l₂ lines intersect at  each other

To Prove :-

The centre of a circle touching two intersecting lines lies on the angle bisector of the lines.

Proof ;-

Assume that O touches l₁ and l₂ at M and N , then we get as ,

⇒ OM = ON ( As it is the radius of the circle )

- Therefore ,From the centre  ''O'' of the circle , it  has equal distance from l₁ & l₂.

Now , In Δ OPM & OPN ,

⇒ OM = ON ( Radius of the circle )

⇒Angle - OMP = Angle -ONP ( As , Radius is perpendicular to its tangent )

⇒ OP = OP ( Common sides )

Therefore , Δ OPM = ΔOPN ( S S S congruence rule )

By C.P.C.T ,

⇒ Angle MPO = Angle NPO

So , l bisects angle MPN.

Threfore , O lies on the bisector of the angle between l₁ & l₂ .

Hence , we prove that ⇒ the centre of a circle touching two intersecting lines lies on the angle bisector of the lines.

Answer 2

Answer:

Step-by-step explanation:

i) Let the two lines l₁ and l₂ intersect at point P. And the circle touches the two lines at A and B respectively. Join center O of the circle with A and B respectively. Also join OP.  

ii) At the point of contact radius and tangent are perpendicular.  

So, <OAP = <OBP = 90 deg  

OP = OP [Common]  

OA = OB [Radii of the same circle]  

Hence ΔOAP ≅ ΔOBP [RHS Congruence axiom]  

So <OPA = <OPB [Corresponding parts of congruence triangles are equal]  

Hence OP is the bisector of angle APB.  

Thus it is proved that O lies on the bisector of the angle formed by the intersecting lines.