Two equal chords \(ST\) and \(UV\) of a circle when produced intersect at a point \(P\). Prove that \(PT= PV\).
Proof:
Given: Two equal chords \(ST\) and \(UV\) of a circle intersecting at a point \(P\).
To prove: \(PT = PV\)
Join \(OP\), draw \(OL ⊥ ST \) and .
Here, \(ST =UV\).
Therefore, [Equal chords are equidistant from the centre]
In \(△OLP\) and \(△OMP\),
\(OL = OM\)
\(∠OLP=∠OMP\) [right angle] and
\(OP =OP\) [common side]
Therefore, \((△OLP≅△OMP)\) [RHS congruence rule]
Now, \(LP = MP\) [by CPCT] .....(1)
Now, \(ST = UV\).
\(\frac{1}{2}ST=\frac{1}{2}UV\)
\(TL=VM\) .....(2) [perpendicular draw from centre to the circle bisects the chord ]
On subtracting (2) from (1) , we get
.
Hence, proved.