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.
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,
In△OPT and △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(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)
Similarly,
In △PRT and △QRT
RT = RT (Common side)
∠OTP = ∠OTQ (from a)
TP = TQ (Tangents drawn from an external point to a circle are equal in length)
∴△PRT ≅ △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)
Also, ∠PRT = ∠QRT (by CPCT)
Now , we know
∠PRT + ∠QRT = 180∘ (Linear pair of angles)
⇒2∠QRT = 180∘ ⇒∠QRT = 90∘=∠PRT→(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.