Question

AB and CD are equal chords of a circle whose centre is O. When produced, these chords meet at E. Then AE = CE.

• A. True
• B. False

Answer 1

The correct option is A

True

From O draw OP$\perp$AB and OQ$\perp$CD. Join OE.

Given, AB = CD.

Since equal chords of a circle are equidistant from the centre, OP = OQ.

Now, in right triangles OPE and OQE,

OE = OE (common)

OP = OQ (proved above)

$⟹△OPE\cong △OQE$ (R.H.S.)

$\therefore$ PE = QE (C.P.C.T.)

$⟹PE-\frac{1}{2}AB=QE-\frac{1}{2}CD$ ($becauseAB=CD$(given))

$⟹PE-PB=QE-QD$

$⟹EB=ED$

$⟹BE+AB=ED+CD(\because AB=CD)$

$⟹AE=CE$

Hence the given statemnet is true.