When two chords of equal length in a circle intersect each other. Show that they are divided into equal segments.
Explanation:
Now, Draw \(OM⊥ WX\), \(ON⊥ YZ\) and join \(OE\), where \(O\) is the centre of circle.
In \(△OME\) and \(△ONE\),
\(OM = \) []
\(OE = \) [common side]
Also, \(∠OME=ONE=\)\(^°\)
Therefore, \(△OME≅△ONE\) []
Now, \(EM=\) [CPCT] .....(1)
Also, \(WX = YZ\).
On dividing both sides by \(2\),
\(\frac{WX}{2}=\frac{YZ}{2}\)
Therefore, \(WM=\) .....(2) [Perpendicular drawn from the centre of a circle to a chord, bisects it]
Now, adding (1) and (2) we get,
\(EM + WM = EN + YN\)
Therefore, \(WE=\) .....(3)
Now, \(WX =YZ \).
On subtracting \(WE\) from both the sides,
\(WX − WE = YZ− WE\)
\(XE = YZ − YE\) [from (3)]
That is, \(XE =\)
Hence, proved.