Solveeit Logo

Question

Question: From a point T outside a circle of centre O, tangents TP and TQ are drawn to the circle . Prove that...

From a point T outside a circle of centre O, tangents TP and TQ are drawn to the circle . Prove that OT is the right bisector of line segment PQ.

Explanation

Solution

Start by drawing the diagram and take triangles OPT and OQT , check for congruency and similarly triangles PRT and QRT , and apply CPCT for both of these congruent triangles . Take the sum of angle PRT and QRT as 180 and solve using the relation found.

Complete step-by-step answer :

In the above diagram TP and TQ are tangents drawn from an external point to the circle. O is the centre of the circle.
Consider OT intersecting PQ at point R.
Now,
InOPT and OQT\vartriangle {\text{OPT and }}\vartriangle {\text{OQT}}
OT = OT (common side)
TP = TQ (Tangents drawn from an external point to a circle are equal in length)
OP = OQ (Radius of the circle)
OPT  OQT\therefore \vartriangle {\text{OPT }} \cong {\text{ }}\vartriangle {\text{OQT}}(by SSS congruency)
[SSS Postulate states that if two triangles have all the corresponding sides equal in length they are said to be congruent.]
By applying CPCT (Corresponding parts of congruent triangles)
OTP = OTQ(a)\angle {\text{OTP = }}\angle {\text{OTQ}} \to ({\text{a)}}
Similarly,
In PRT and QRT\vartriangle {\text{PRT and }}\vartriangle {\text{QRT}}
RT = RT (Common side)
OTP = OTQ\angle {\text{OTP = }}\angle {\text{OTQ}} (from a)
TP = TQ (Tangents drawn from an external point to a circle are equal in length)
PRT  QRT\therefore \vartriangle {\text{PRT }} \cong {\text{ }}\vartriangle {\text{QRT}}(by SAS congruency)
[SAS postulate states that if two corresponding sides and an included angle of two triangles are equal then they are congruent.]
By applying CPCT (Corresponding parts of congruent triangles)
PR = QR(b){\text{PR = QR}} \to {\text{(b)}}
Also, PRT = QRT\angle {\text{PRT = }}\angle {\text{QRT}} (by CPCT)
Now , we know
PRT + QRT = 180 \angle {\text{PRT + }}\angle {\text{QRT = 18}}{0^ \circ }{\text{ }} (Linear pair of angles)
2QRT = 180 QRT = 90=PRT(c)  \Rightarrow 2\angle {\text{QRT = 18}}{0^ \circ } \\\ \Rightarrow \angle {\text{QRT = 9}}{0^ \circ } = \angle {\text{PRT}} \to (c) \\\
From relation b and c , we can say that
OT is the right bisector of the line segment PQ.
Hence proved.

Note : Properties of congruent triangles along with all the postulates for congruence must be well known , also properties of the circle are must in order to solve such similar questions. Attention is to be given while selecting the triangles and their corresponding side and angles.