An isosceles triangle \(LMN\) is inscribed in a circle, with \(LM = LN\). Justify that the altitude from \(L\) to \(MN\) passes through the centre of the circle.
Proof:
Let the altitude from \(L\) meet \(MN\) at \(O\).
Consider \(\Delta LOM\) and \(\Delta LON\).
The sides \(LM = LN\) []
\(LO = LO\) []
\(\angle LOM\) \(=\) \(LON\) \(=\) \(^{\circ}\)
Therefore, \(\Delta LOM ≅ \Delta LON\). [By ]
Thus, \(OM = ON\) [By ]
This implies that, \(O\) is the of the side \(MN\).
By the theorem:
Therefore, the altitude from \(L\) to \(MN\) passes through the centre of the circle.
Hence, proved.