Question

Two equal chords AB and CD of a circle with centre O, when produced meet at a point E. Prove that BE = DE and AE = CE.

Answer 1

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Given: Two equal chords AB and CD are intersecting at a point E.

To Prove: BE = DE and AE = CE

Construction: Join OE, Draw OL$\perp$ AB and OM$\perp$ CD

Proof: AB = CD (Given)

In $△$ OLE and $△$OME,

OL = OM (equal chords are at equal distance from the centre)

$\angle$ OLE = $\angle$ OME (Each equal to 90^o)

OE = OE (common side)

∴ $△$OLE$\cong$$△$OLE (By RHS Theorem of congruence)

LE = ME …..(1) (C.P.C.T)

AB = CD (Given)

Now, $\frac{1}{2}$AB = $\frac{1}{2}$CD ⇒ BL = DM ……(2)

Subtracting (2) from (1), we get

LE–BL = ML-DM

BE=DE

Again, AB=CD (given) and BE=DE ......(3)

AB+BE = CD+DE (from (3))

AE=CE

Hence, BE=DE and AE=CE

AB + BE = CD + DE

AE = CE

Hence, BE = DE and AE = CE.