Question

If two equal chords of a circle intersect within the circle, prove that the segments of one chord are equal to corresponding segments of the other chord.

Answer 1

Step-1 Given and construction:

Let two equal chords $AB$ & $CD$ intersect at $E.$

Draw$OM$ perpendicular $AB,ON$perpendicular$CD$and join$OE.$

Step-2 Calculation to prove $AE=ED$ and $CE=EB$

Because perpendicular to the center bisects the chord.

$$\begin{gathered} AM=MB=\frac{1}{2}AB \\ CN=ND=\frac{1}{2}CD \\ AM=MB=CN=ND...\left(1\right) \end{gathered}$$

Now, In$∆OME$ and$∆ONE$

$\angle M=\angle N$ [$90°$both]

$OE=OE$ [Common]

$ON=OM$ [equal chords are at equal distance from the center]

$\therefore ∆OME\cong ∆ONE$ (RHS criteria)

Therefore $ME=EN$$...\left(2\right)$

From$\left(1\right)$and$\left(2\right)$

$$\begin{gathered} AM+ME=ND+AN \\ \Rightarrow AE=ED \end{gathered}$$

$$\begin{gathered} MB-ME=CN-EN \\ \Rightarrow EB=EC \end{gathered}$$

Hence it is proved that $AE=ED$ and $CE=EB$ .