AB and CD are two equal chords of a circle whose centre is O. When produced these chords meet at E. Prove that EB=ED

Given: AB and CD are two equal chords of a circle whose centre is O. AB and CD , when produced meet at E.

To Prove: EB=ED and EA=EC

Construction: Join OE. Draw OP⊥ AB and OQ⊥ CD 

Proof : Since equal chords are equidistant from the centre.

∴AB = CD 
⇒ OP = OQ .........(i)

Also, ∠OPE = ∠OQE=90^o 
ΔOPE  ΔOQE [By RHS congruence criterion]

PE = QE[By Cpct]........... (ii) 

Since OP⊥AB

Since OQ⊥CD

Now, from (iii)  & (iv), we get,

AP=PB=CQ=QD (Since AB=CD)........(v)

On adding (v) & (ii), we get,
 

AP+PE=CQ+QE

⇒AE=CE.........(vi)

and  

PE-AP=EQ-CQ

⇒PE-PB=EQ-ED  [Using(v)]

⇒EB=DE...............(vii)

(vi)  and (vii) prove the result.