Are you allowed to use trigonometric relations?
The triangles BPM and CPB are similar, this allows to find coordinates of the point P. Together with D and N, you can calculate the angle.
I have spent hours trying to solve this maths problem but unfortunately I just cannot figure out how this can be solved.
The question reads: The figure shows a square ABCD. M and N are points on AB and BC respectively such that BM=BN. Take a point P on MC such that BP is perpendicular to MC. Prove that PD is perpendicular to PN.
Can you please elaborate further? I fail to see why BPM and CPD are similar triangles, and even if this is the case I don't know how to obtain the coordinate of P.
And how do I work out the angle once I know the coordinate of P?
Both triangles have a 90°-angle and the angle BMC and MCB add up to 90°, therefore the angle BMC is equal to the angle CBP.
Scalar product of the corresponding vectors. But I think this answes my question as "no".And how do I work out the angle once I know the coordinate of P?
Start by showing that the triangles DCP and NBP are similar.
In the triangles and we have
(both ratios are the tangents of the equal angles and ).
and together show that the triangles DCP and NBP are similar.
From that it follows that the angles and are equal in which case angles and sum to degrees.
It then follows that since the interior angles of a quadrilateral sum to degrees, that, (from the quadrilateral ) that the angles and sum to , in which case degrees.