Answer :
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.
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