Question

In the figure, 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 and EA = EC.

Please provide complete steps and justifications.

Answer 1

see the picture for the solution

Answer 2

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

RTP :i) EB= ED

ii) EA = EC

consruction: from O draw OP ⊥AB and OQ ⊥ CD.join O to E.

proof : AB = CD [ given]
OP= OQ [ ∵ equal chords of a circle are equidistant from the centre]

In ΔOPE and ΔOQE

OE = OE [hypotenuse]
OP= OQ [side]
∴ΔOPE ≡ ΔOQE
[RHS rule }

PE = QE [ cpct - corresponding parts of congruent triangles]

PE -AB/2 = QE - CD/2
[ since AB= CD given]
PE -PB = QE -QD
EB = ED.---(1)

BE +AB = ED +CD [ since AB = CD]

EA = EC----(2)
if u want pic i will send it