Advertisements
Advertisements
Question
Given two equal chords AB and CD of a circle with center O, intersecting each other at point P.
Prove that:
(i) AP = CP
(ii) BP = DP
Solution
In ΔOMP and ΔONP,
OP = OP ...( common sides )
∠OMP = ∠ONP ...( both are right angles )
OM = OM ...( side both the chords are equal, so the distance of the chords from the centre are also equal )
ΔOMP ≅ ΔONP ...( RHS congruence criterion )
⇒ MP = PN ...(c.p.c.t ) ....( a )
(i) Since AB = CD ...( given )
⇒ AM = CN ...( drawn from the centre to the chord bisects the chord )
⇒ AM + MP = CN + NP .....( from a )
⇒ AP = CP ....( b )
(ii) Since AB = CD
⇒ AP + BP = CP + DP
⇒ BP = DP ....( from b )
Hence proved.
APPEARS IN
RELATED QUESTIONS
AB and CD are two equal chords of a circle with centre O which intersect each other at right
angle at point P. If OM⊥ AB and ON ⊥ CD; show that OMPN is a square.
In the following figure, O is the centre of the circle. Find the values of a, b and c.
In the following figure, O is the centre of the circle. Find the values of a, b, c and d.
In the following figure, O is the centre of the circle. Find the values of a, b, c and d.
In the following figure, O is centre of the circle and ΔABC is equilateral.
Find:
- ∠ADB,
- ∠AEB.
In the given figure, I is the incentre of ΔABC. BI when produced meets the circumcircle of ΔABC at D. ∠BAC = 55° and ∠ACB = 65°; calculate:
- ∠DCA,
- ∠DAC,
- ∠DCI,
- ∠AIC.
In following fig., chords PQ and RS of a circle intersect at T. If RS = 18cm, ST = 6cm and PT = 18cm, find the length of TQ.
In the figure, chords AE and BC intersect each other at point D. If AD = BD, show that AE = BC.
Two chords AB, CD of lengths 16 cm and 30 cm, are parallel. If the distance between AB and CD is 23 cm. Find the radius of the circle.
AB and CD are two chords of a circle such that AB = 6 cm, CD = 12 cm and AB || CD. If the distance between AB and CD is 3 cm, find the radius of the circle.
