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 $90^\circ ]$ 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 $
$\Rightarrow PB = PD$ 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 $90^\circ ]$ 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 $
$\Rightarrow PB = PD$ Hence proved.
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