Question

Question 1
If two equal chords of a circle intersect, prove that the parts of one chord are separately equal to the parts of the other chord.

Answer 1

Given AB and CD are two equal chords of a circle, let they meet at a point E.
To Prove that AE = CE and BE = DE
Construction : Draw OM $\perp$ AB and ON $\perp$ CD and join OE where O is the centre of the circle.

Proof
In $\Delta OMEand\Delta ONE$
OM = ON [equal chords are equidistant from the centre]
OE = OE [Common side]
And $\angle OME=\angle ONE\left[each{90}^{\circ }\right]$
$\therefore \Delta OME\cong \Delta ONE$ [by RHS congruence rule]
$\Rightarrow$ EM = EN [by CPCT] …… (i)

Now, AB = CD
On dividing both sides by 2, we get
$\frac{AB}{2}=\frac{CD}{2}$
AM=CN............(ii)
[since, perpendicular drawn from centre of circle to chord bisects the chord i.e., AM = MB and CN = ND]
On adding Eqs. (i) and (ii), we get
AM + EM = CN + EN
AE = CE............(iii)

We have AB = CD
On subtracting both sides by AE, we get
AB – AE = CD – AE
BE = CD – CE [from Eq. (iii)]
BE = DE, hence proved.