Two equal chords AB and CD of a circle with centre O , when produced meet at point E , as shown in figure. Prove that BE = DE and AE= CE
Answers
Answer:
Given : Two equal chords AB and CD intersecting at a point E.
To prove : BE = DE and AE = CE
Construction : Join OE. Draw OL ⊥ AB and OM ⊥ CD.
Proof : We have
AB = CD
⇒ OL = OM [∵ Equal chords are equidistant from the centre ]
In triangles OLE and OME, we have
OL = OM
∠OLE = ∠OME [Each equal to 90˚]
And OE = OE [Common]
So, by SAS – criterion of congruences
△OLE ≅ △OME
⇒ LE = ME ……..(i)
Now, AB = CD
⇒ ½ AB = ½ CD ⇒ BL = DM …….(ii)
Subtracting (ii) from (i), we get
LE – BL = ME – DM
⇒ BE = DE
Again, AB = CD and BE = DE
⇒ AB + BE = CD + DE
⇒ AE = CE
Hence, BE = DE and AE = CE
Hence proved.
Step-by-step explanation:
Given : Two equal chords AB and CD intersecting at a point E.
To prove : BE = DE and AE = CE
Construction : Join OE. Draw OL ⊥ AB and OM ⊥ CD.
Proof :
In △OLE and △OME, we have
OL = OM [Equal chords are equidistant from the center ]
∠OLE = ∠OME [Each equal to 90˚]
OE = OE [Common]
So,
△OLE ≅ △OME [SAS congruence rule]
⇒ LE = ME (i)[CPCT]
Now, AB = CD
⇒ ½ AB = ½ CD
⇒ BL = DM (ii)
Subtracting (ii) from (i), we get
LE – BL = ME – DM
⇒ BE = DE
Again, AB = CD and BE = DE
⇒ AB + BE = CD + DE
⇒ AE = CE
Hence, BE = DE and AE = CE
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