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Question
Two equal chords AB and CD of a circle with center O, when produced meet at a point E, as shown in Fig. Prove that BE = DE and AE = CE.
Solution
Given: Two equal chords AB and CD intersecting at a point E.
To prove: BE = BE 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
Δ OLE ≅ ΔOME
⇒ LE = ME ....(i)
Now, AB = CD
⇒ `1/2"AB" = 1/2"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.
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