# Prove that the angle between two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line segment joining the points of contact at the centre.

This figure will surely help u to understand easily ......

To prove==>  Angle BPA + Angle BOA = 180 ......

Angle PBO = 900 [As angle between the radius and the tangent to the circle is 900]

Similarly angle PAO = 90.......

Angles (PAO + PBO + BOA + BPA) = 3600 ( sum of angles of a quadrilateral)

----->  900 + 900 + angle BOA + angle BPA = 3600 (<PAO=90 and <PBO=90 proved)

----->  Angle BOA + Angle BPA = 3600 - 1800....

----->  Angle BOA + Angle BPA = 1800

!!!!....Hence proved.....Hope this helps....!!!!

• 161
thanks

• -7
figure is not visible...

• -6
YAYAYA
• -15
Prove that the parallelogram circumscribing a circle is a rhombus
• -1
This is correct

• 0
What you will be doing in class while your teacher is teaching.
• -6
Then go visit theopthalmologist...!!!
• -3
What are you looking for?