Question 11 Mark
Prove that the lengths drawn at the diameter of a circle are parallel.
Answer
Given: CD and EF are the tangent at endpoint A and B of the diameter AB of a circle with centre O. To trove: CD || EF Proof: Since CD is the tangent to the circle at the point A,$\angle\text{BAD}=90^\circ$
Since CD is the tangent to the circle at the point B,$\angle\text{ABE}=90^\circ$
Thus, $\angle\text{BAD}=\angle\text{ABE}=90^\circ$ But these are alternate interior angles. ⇒ CD || EF Hence proved.
View full question & answer→ Given: CD and EF are the tangent at endpoint A and B of the diameter AB of a circle with centre O. To trove: CD || EF Proof: Since CD is the tangent to the circle at the point A,$\angle\text{BAD}=90^\circ$Since CD is the tangent to the circle at the point B,$\angle\text{ABE}=90^\circ$
Thus, $\angle\text{BAD}=\angle\text{ABE}=90^\circ$ But these are alternate interior angles. ⇒ CD || EF Hence proved.
Given: Two tangent RA and RA are drawn from a point A to a circle with centre O. To prove: $\angle\text{RAB}=\angle\text{RBA}$ Proof: We know that tangent drawn from an external point to the circle are equal. So, in $\triangle\text{RAB},$ RA = RB$\Rightarrow\angle\text{RAB}=\angle\text{RBA}$ ....(side opposite equal angles are equal)
Two tangent AP and AQ are drawn from a point A to a circle with centre O.
Given: Two tangent AP and AQ are drawn from a point A to a circle with centre O. To prove: AP = AQ Construction: Join OP, OQ and OA. Proof: Since AP and AQ are the tangents to the circle, $\text{OP}\perp\text{AP}$ and $\text{OQ}\perp\text{AQ}$ In $\triangle\text{OPA}$ and $\triangle\text{OQA},$$\angle\text{OPA}=\angle\text{OQA}=90^\circ$
Given: A parallelogram ABCD circumscribes a circle with centre O, To prove: AB = BC = CD = AD Proof: We know that tangents drawn from an external point to the circle are equal.$\therefore$ AP = AS ....(tangent from A)