From an external point two tangents are drawn to a circle. Prove that the line joining the external point to the centre of the circle bisects the angle between the two tangents as well as the angle between the two normals at the point of contact.
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Answers
Answer:
Step-by-step explanation:
Draw a circle with center OO and take a external point P. PA and PB are the tangents.
As radius of the circle is perpendicular to the tangent.
OA \bot PAOA⊥PA
Similarly OB\bot PBOB⊥PB
\angle OBP=90^o∠OBP=90
o
\angle OAP = 90^o∠OAP=90
o
In Quadrilateral OAPBOAPB, sum of all interior angles = 360^o=360
o
\Rightarrow \angle OAP+\angle OBP+\angle BOA + \angle APB =360^o⇒∠OAP+∠OBP+∠BOA+∠APB=360
o
\Rightarrow 90^o+90^o+\angle BOA+\angle APB = 360^o⇒90
o
+90
o
+∠BOA+∠APB=360
o
\angle BOA + \angle APB = 180^o∠BOA+∠APB=180
o
It proves the angle between the two tangents drawn from an external point to a circle supplementary to the angle subtented by the line segment
solution
Step-by-step explanation:
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