An isosceles triangle \(PQR\) is inscribed in a circle, with \(PQ = PR\). Deduce that the altitude from \(P\) to \(QR\) passes through the centre of the circle.
Proof:
Let the altitude from \(P\) meet \(QR\) at \(S\).
Consider \(\Delta PSQ\) and \(\Delta PSR\).
The sides \(PQ = PR\) []
\(PS = PS\) []
\(\angle PSQ\) \(=\) \(\angle PSR\) \(=\) \(^{\circ}\)
Therefore, \(\Delta PSQ ≅ \Delta PSR\). [By ]
Thus, \(SQ = SR\) [By ]
This implies that, \(S\) is the of the side \(QR\).
By the theorem:
Therefore, the altitude from \(P\) to \(QR\) passes through the centre of the circle.
Hence, proved.