Question
Two equal chords AB and CD of a circle when produced intersect at a point P. Prove that PB = PD
✓
Answer
Given: Two equal chords AB and CD of a circle intersecting at a point P.
To prove: PB = PD
Construction: Join OP, draw $\text{OL}\bot\text{AB}$ and $\text{OM}\bot\text{CD}$
Proof: We have, AB = CD
$\Rightarrow\text{OL}=\text{OM}$ [equal chords are equidistant from the centre]
In $\triangle\text{OLP}$ and $\triangle\text{OMP}\ \ \text{OL}=\text{OM}$ [proved above]
$\angle\text{OLP}=\angle\text{OMP}$ [each 90°]
and OP = OP [common side]
$\therefore\triangle\text{OLP}=\triangle\text{OMP}$ [by RHS congruence rule]
$\Rightarrow\text{LP}=\text{MP}$ [by CPCT] ...(i)
Now, AB = CD
$\Rightarrow\frac{1}{2}(\text{AB})=\frac{1}{2}(\text{CD})$ [dividing both sides by 2]
$\Rightarrow\text{BL}=\text{DM}\ \ ...(\text{ii})$
[perpendicular drawn from centre to the circle bisects the chord i.e., AL = LB and CM = MD]
On subtracting Eq. (ii) from Eq. (ii) we get
LP - BL = MP - DM ⇒ PB = PD
Hence proved.
To prove: PB = PD
Construction: Join OP, draw $\text{OL}\bot\text{AB}$ and $\text{OM}\bot\text{CD}$
Proof: We have, AB = CD
$\Rightarrow\text{OL}=\text{OM}$ [equal chords are equidistant from the centre]
In $\triangle\text{OLP}$ and $\triangle\text{OMP}\ \ \text{OL}=\text{OM}$ [proved above]
$\angle\text{OLP}=\angle\text{OMP}$ [each 90°]
and OP = OP [common side]
$\therefore\triangle\text{OLP}=\triangle\text{OMP}$ [by RHS congruence rule]
$\Rightarrow\text{LP}=\text{MP}$ [by CPCT] ...(i)
Now, AB = CD
$\Rightarrow\frac{1}{2}(\text{AB})=\frac{1}{2}(\text{CD})$ [dividing both sides by 2]
$\Rightarrow\text{BL}=\text{DM}\ \ ...(\text{ii})$
[perpendicular drawn from centre to the circle bisects the chord i.e., AL = LB and CM = MD]
On subtracting Eq. (ii) from Eq. (ii) we get
LP - BL = MP - DM ⇒ PB = PD
Hence proved.
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