Two Equal chords AB and CD when produced meet at point P prove that pa pc

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Question 11 Two equal chords AB and CD of a circle when produced intersect at a point P. Prove that PB = PD.

Solution

Given two equal chords $A B$ and $C D$ of a circle interesting at a point $P .$

To prove that $P B = P D .$
Construction : Join $O P,$ draw $O L$ $⊥$ $A B$ and $O M$ $⊥$ $C D .$
Proof: We have, $A B = C D$
OL = OM [equal chords are equidistant from the centre]
In $Δ O L P$ and $Δ O M P$ , we have
OL = OM [proved above]
$∠ O L P = ∠ O M P$ [each $90^{\circ}]$
And, $O P = O P$ [common side]
$∴ Δ O L P ≅ Δ O M P$ [by RHS congruence rule]
$L P = M P$ [by CPCT] ……(i)
Now, $A B = C D$

$\Rightarrow \frac{1}{2} (A B) = \frac{1}{2} (C D)$ [dividing both sides by 2]
$\Rightarrow$ $B L = D M$ ...............(ii)
[perpendicular draw from centre to the circle bisects the chord i.e., $A L = L B$ and $C M = M D$ ]
On subtracting Eq. (ii) from Eq. (i), we get
$L P - B L = M P - D M$
$\Rightarrow$ $P B = P D .$ hence proved.

Mathematics
NCERT Exemplar
Standard IX

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