If two equal chords of a circle intersect within the circle, prove that the segments of one chord are equal to corresponding segments of the other chord.
Answers
Solution :-
Let PQ and RS be two equal chords of a given circle and they are intersecting each other at point T .
Draw perpendicular OV and OU on these chords.
In △ OVT and △ OUT ,
OV = OU ( Equal chords of a circle are equidistant from the centre )
∠ OVT = ∠ OUT ( Each 90°)
OT = OT ( common side)
Therefore, △ OVT ≅ △ OUT ( RHS congruence rule)
VT = UT ( By CPCT) —— (1)
It is given that,
PQ = RS —— (2)
=> ½ PQ = ½ RS
=> PV = PU —— (3)
On adding equation (1) and (3), we obtain
PV + VT = RU + UT
=> PT = RT —— (4)
On subtracting equation (4) from equation (2) , we obtain
PQ ‐ PT = RS - RT
=> QT = ST —— (5)
Equation (4) and (5) indicate that the corresponding segments of chords PQ and RS are congruent to each other.
Drop a perpendicular from O to both chords AB and CD
In △OMP and △ONP
As chords are equal, perpendicular from centre would also be equal.
OM=ON
OP is common.
∠OMP=∠ONP=90o
△OMP ≅ △ONP (RHS Congruence)
PM=PN (1)
AM=BM (Perpendicular from centre bisects the chord)
Similarly ,CN=DN
As AB=CD
AB−AM=CD−DN
BM=CN (2)
From (1) and (2)
BM−PM=CN−PN
PB=PC
AM=DN (Half the length of equal chords are equal)
AM+PM=DN+PN
AP=PD
Therefore , PB=PC and AP=PD is proved.