**[RMO-1999]**: *Let ABCD be a square and \(M\) & \(N\) points on sides \(AB\) & \(BC\) respectively such that \(\angle MDN = 45^o\). If \(R\) is the mid point of \(MN\), show that \(RP = RQ\), where \(P\) & \(Q\) are the points of intersection of \(AC\) with the lines \(MD\) & \(ND\)*

**Solution**:

Join \(PN\) and \(MQ\). Since \(\angle PDN = \angle PCN\), PDCN is a cyclic quadrilateral. It can now be inferred that \(\angle PND = \angle PCD = 45^o\).

Now look at \(\triangle PDN. Since \angle PDN = \angle PND = 45^o, => \angle NPD = \angle NPM = 90^o\). Similarly, \(\angle MQD = \angle MQN = 90^o\).

Since \(\angle NPM = \angle MQN = 90^o\), \(MPQN\) is also a cyclic quadrilateral. Furthermore, \(MN\) becomes the diameter of the circumcircle of \(MPQN\) since \(MN\) project \(90^o\) on the sector. Since \(R\) is the midpoint of \(MN, => R\) is the centre of the circumcircle of \(MPQN\). Thus, \(RP = RQ\)