???!?!!?!!?????!!!!!????????solve it plz
Attachments:
Answers
Answered by
1
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.
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.
Similar questions