Two chords of a circle intersect and each makes the same angle with the diameter passing through their point of intersection. Show that these chords are equal in length
Proof:
Draw perpendiculars \(OL\) and \(OM\) on chords \(AB\) and \(CD\), respectively.
Now in \(\bigtriangleup LOE\) \(∠ LOE = 180^° – 90^° – ∠ LEO\) [Angle sum property of a triangle]
\(= 90^° – ∠ LEO\)
\(= 90^° – ∠\)
\(= 90^° – ∠ \)
\(= 90^° – ∠ MEO = ∠ \) [Angle sum property]
In triangles \(\bigtriangleup OLE\) and \(\bigtriangleup OME\),
\(∠ LEO = ∠ \)
\(∠ LOE = ∠ \) (Proved above)
\(EO = EO\) (Common)
Therefore, \(∆ OLE ≅ ∆ OME\) ()
This gives \(OL = OM\) ()
So, \(AB = CD\)