Prove that the tangents drawn at the ends of a chord of a circle make equal angles with the chord.
(N/A) Let $PR$ be a chord of a circle. Tangents are drawn at the points $P$ and $R$. Let these tangents meet at a point $T$ outside the circle.
In $\triangle TPR$,$TP = TR$ (tangents drawn from an external point to a circle are equal in length).
Since $TP = TR$,the angles opposite to these sides are equal,i.e.,$\angle TPR = \angle TRP$.
These angles are the angles made by the tangents with the chord $PR$.
Hence,the tangents drawn at the ends of a chord of a circle make equal angles with the chord.
In $\triangle TPR$,$TP = TR$ (tangents drawn from an external point to a circle are equal in length).
Since $TP = TR$,the angles opposite to these sides are equal,i.e.,$\angle TPR = \angle TRP$.
These angles are the angles made by the tangents with the chord $PR$.
Hence,the tangents drawn at the ends of a chord of a circle make equal angles with the chord.
Explore More
Similar Questions
If $AB$ is a chord of a circle with centre $O$,$AOC$ is a diameter and $AT$ is the tangent at $A$ as shown in the figure. Prove that $\angle BAT = \angle ACB$.
Medium
View SolutionIn $Fig.$,$PQ$ is a chord of a circle and $PT$ is the tangent at $P$ such that $\angle QPT = 60^{\circ}$. Then $\angle PRQ$ is equal to (in $^{\circ}$)
Easy
View SolutionIn the figure,tangents $PQ$ and $PR$ are drawn to a circle such that $\angle RPQ = 30^{\circ}$. $A$ chord $RS$ is drawn parallel to the tangent $PQ$. Find the $\angle RQS$. (in $^{\circ}$)
Difficult
View SolutionIn the figure,from an external point $P$,a tangent $PT$ and a line segment $PAB$ are drawn to a circle with centre $O$. $ON$ is perpendicular to the chord $AB$. Prove that:
$(i) \quad PA \cdot PB = PN^2 - AN^2$
$(ii) \quad PN^2 - AN^2 = OP^2 - OT^2$
$(iii) \quad PA \cdot PB = PT^2$
$(i) \quad PA \cdot PB = PN^2 - AN^2$
$(ii) \quad PN^2 - AN^2 = OP^2 - OT^2$
$(iii) \quad PA \cdot PB = PT^2$
Difficult
View Solution$\overline{AB}$ is a chord of $\odot(O, 13)$ such that $AB = 24$. Tangents at $A$ and $B$ to the circle intersect at $P$. Find $PA$.
Difficult
View SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo