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. [3 MARKS]

Answer 1

Concept: 1 Mark
Steps: 1 Mark
Answer: 1 Mark

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

To prove :EB = ED

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

Proof :

$AB=CD$ Given

$OP=OQ$ .... (1) [ equal chords of a circle are equidistant from the center]

In $\Delta OPE$ and $\Delta OQE$

$OE=OE$ [ Common side]

$OP=OQ$ [ From (1) ]

$\angle OPE=\angle OQE={90}^{\circ }$

$\therefore \Delta OPE\cong OQE$ By R.H.S

$\therefore PE=QE$ [By C.P.C.T]

$PE-\frac{AB}{2}=QE-\frac{CD}{2}$ [ $\because$ AB=CD ( Given)]

$PE-PB=QE-QD$

$EB=ED$

Hence proved