Two equal chords AB and CD of a circle when produced intersect at a point P. Prove that PB = PD.
Solution:
Given, AB and CD are two equal chords of a circle.
The chords AB and CD intersect at a point P.
We have to prove that PB = PD
Join OP
Draw OL perpendicular to AB and OM perpendicular to CD
We know that the two equal chords are equidistant from the centre of a circle.
Now, AB = CD
OL = OM
Considering triangles OLP and OMP,
OL = OM
∠OLP = ∠OMP
Common side = OP
By RHS criterion, the triangles OLP and OMP are similar.
By CPCTC,
LP = MP --------------------------- (1)
We know that the perpendicular drawn from centre to the circle bisects the chord.
AL = LB
CM = MD
Now, AB = CD
Dividing by 2 on both sides,
1/2 AB = 1/2 CD
So, BL = DM ---------------------- (2)
Subtracting (1) and (2),
LP - BL = MP - DM
Therefore, PB = PD
✦ Try This: In a circle of radius 21 cm, an arc subtends an angle of 60° at the Centre. Find the length of the arc.
☛ Also Check: NCERT Solutions for Class 9 Maths Chapter 10
NCERT Exemplar Class 9 Maths Exercise 10.4 Problem 11
Two equal chords AB and CD of a circle when produced intersect at a point P. Prove that PB = PD.
Summary:
Two equal chords AB and CD of a circle when produced intersect at a point P. It is proven that PB = PD
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