Question

In fig. the centre of the circle is O. PQ and RS are two equal chords of the circle which , when produced , meet at T outside the circle . Prove that (a) TP = TR (b) TQ = TS.

Answer 1

Given PQ = RS 

To prove : TP = TR and TQ = TS 

Construction : Draw ON ⊥ PQ and OM ⊥ RS

Proof :  Since equal vhords are equidistance from the circle therefore 

PQ = RS  ⇒ ON = OM       ....(1)

Also perpendicular drawn from the centre bisects the chord.

So, PN = NQ = `1/2` "PQ"  and RM = MS = `1/2` "RS"

But PQ = RS , we get

PN = RM         ......(2)

And , NQ = MS      ....(3)

Now in Δ TMO and Δ TNO ,

TO = TO      .....(common)

MO = NO     .....(BY (1))

∠ TMO = ∠ TNO   ...(Each 90 degrees)

Therefore , Δ TMO ≅ Δ TNO ,   ..(By RHS)

⇒ TN = TM    ...(by CPCT)  ...(4)

Substracting ,(2) from (4) , we get

TN - PN = TM - RM

⇒ TP = TR

Adding (3) and (4) , we get

TN + NQ = TM + MS

⇒ TQ = TS

Adding (3) and (4) , we get

TN + NQ = TM + MS

⇒ TQ = TS

Hence proved.