Question

AB and CD are equal chords of a circle whose centre is O. When produced, these chords meet at E. Prove that EB = ED. (figure 4)

Answer 1

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=90o 
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

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=90o 
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

Answer 2

ANSWER:-

Given:- AB=CD

To prove:- EB=ED

Construction:- Join OE, draw OP ⊥ AB, OQ ⊥ CD.

Proof:- In Δ's EPO AND EQO,

EO = EO (common)

OP = OQ ( equal chords are equidistant from the centre )

∠EPO = ∠EQO ( Both are 90º)

therefore, ΔEPO ≅ ΔEQO

EP = EQ (CPCT)

As we know that,

AP = CQ ( OP bisects AB and OQ bisects CD )

AP + EP = CQ + EQ

∴EA = EC

EA - AB = EC - CD

∴EB = ED

THANKS!!!