Manipur Board Solutions for Chapter: Circles, Exercise 1: EXERCISE 8.1

Prove that perpendicular at the point of contact to the tangent to a circle passes through the centre.

Prove that opposite sides of the quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.

If the circle touches all the four sides of the quadrilateral $\mathrm{ABCD}$, prove that $\mathrm{AB}+\mathrm{CD}=\mathrm{BC}+\mathrm{AD}$.

$∆\mathrm{ABC}$ is isosceles with $\mathrm{AB}=\mathrm{AC}$. The incircle of the $∆\mathrm{ABC}$ touches $\mathrm{BC}$ at $\mathrm{P}$. Prove that $\mathrm{BP}=\mathrm{CP}$.

Prove that parallelogram circumscribing a circle is a rhombus.

The incircle of $∆\mathrm{ABC}$ touches the sides $\mathrm{BC}, \mathrm{CA} a\mathrm{nd} \mathrm{AB}$ at $\mathrm{D}, \mathrm{E}, \mathrm{F}$. Show that $\mathrm{AF}+\mathrm{BD}+\mathrm{CE}=\mathrm{AE}+\mathrm{BF}+\mathrm{CD}=\frac{1}{2}\left(\mathrm{Perimeter} \mathrm{of} ∆\mathrm{ABC}\right)$.

$\mathrm{PA}$ and $\mathrm{PB}$ are tangents drawn from an external point $\mathrm{P}$ to a circle with centre $\mathrm{O}$. Prove that $\angle \mathrm{OAB}=\frac{1}{2}\angle \mathrm{APB}$.

Prove that the intercept of a tangent between two parallel tangents to a circle subtends a right angle at the centre.