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प्रश्न
In the given Figure, AB and CD are two chords of a circle, intersecting each other at P such that AP = CP. Show that AB= CD.
उत्तर १
If two chords of a circle interest internally then the products of the lengths of segments are equal, then
AP x BP= CP x DP ... ( 1)
But, AP= CP (Given) ....(2)
Then from ( 1) and (2), we have
BP= DP ......(3)
Adding (2) and (3),
AP + BP= CP + DP
⇒ AB = CD
Hence Proved.
उत्तर २
In order to prove the desired result, we shall first prove that ΔPAD ∼ ΔPCB.
In triangles PAD and PCB, we have:
∠ PAD = ∠PCB ...[Angles in the same segment of arc BD]
∠ APD = ∠ CPB ...[Vertically opposite angles]
So, by AAA criterion of similarity, we have
ΔPAD ∼ ΔPCB.
⇒ `"PA"/"PC" = "PD"/"PB"` ....[Corresponding sides of similar triangles are in the same ratio]
⇒ `"AP"/"CP" = "PD"/"PB"`
⇒ 1 = `"PD"/"PB" ...[ ∴ "AP" = "CP", "AP"/"CP" = 1]`
⇒ PB = PD
⇒ AP + PB = AP + PD ....[ Adding AP on both sides ]
⇒ AP + PB = CP + PD ...[ ∵AP = CP ]
⇒ AB = CD.
Hence proved.
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