If two equal chords of a circle intersect each other. Then prove that two parts of one chord are respectively equal to two parts of another chord.

Let AB and CD be two equal chords of a circle which are intersecting at a point E.

Construction:

Draw perpendiculars OF and OG on the chords.

Join OE.

Consider ΔOFE and ΔOGE,

⇒ OF = OG [Equal chords]

⇒∠OFE = ∠OGE [Each 90°]

⇒ OE = OE [Common]

By RHS congruence rule,

⇒ ΔOFE ≅ ΔOGE

By CPCT,

⇒ FE = GE … (1)

Given AB = CD … (2)

⇒ 1/2 AB = 1/2 CD

⇒ AG = CF … (3)

Adding equations (1) and (3),

⇒ AG + GE = CF + FE

⇒ AE = CE … (4)

Subtracting equation (4) from (2),

⇒ AB – AE = CD - CE

⇒ BE = DE … (5)

From (4) and (5),

We can see that two parts of one chord are equal to two parts of another chord.

Hence proved.