An isosceles triangle \(EFG\) is inscribed in a circle, with \(EF = EG\). Demonstrate that the altitude from \(E\) to \(FG\) passes through the centre of the circle.
Proof:
Let the altitude from \(E\) meet \(FG\) at \(H\).
Consider \(\Delta EHF\) and \(\Delta EHG\).
The sides \(EF = EG\) []
\(EH = EH\) []
\(\angle EHF\) \(=\) \(\angle EHG\) \(=\) \(^{\circ}\)
Therefore, \(\Delta EHF ≅ \Delta EHG\). [By ]
Thus, \(HF = HG\) [By ]
This implies that, \(H\) is the of the side \(FG\).
By the theorem:
Therefore, the altitude from \(E\) to \(FG\) passes through the centre of the circle.
Hence, proved.