I encountered the following problem:
The point K is interior to the parallelogram ABCD and is such that the midpoint M of the side
BC is equidistant from points K and D, and the midpoint N of side CD is equidistant from
points K and B. Point P is the midpoint of segment AK. Prove that $\angle{PBK} =\angle{PDK}$
This problem is approachable with synthetic geometry but personally I think that this solution is quite "ugly" and "unnatural" which is why I wanted to resolve this problem with more of a analytic approach
My first attempt was to use polar coordinates, yet I remained uncertain of how to express the equidistant condition. Thus, I found myself in need of assistance and humbly sought guidance. May I be so fortunate as to find the answer I'm searching for.
For the synthetic sol: Through the vertex C of the parallelogram we build a line parallel to the diagonal BD, intersecting AB and AD at points $B_1$ respectively and $D_1$. This creates two new parallelograms and further reasoning solves the problem.
Diagram:
Construct the blue circle with N as center and radius = NB = NK.
Draw NS perpendicular to BK cutting BK at S. Then S is the midpoint of BK. Applying ‘midpoint theorem’ to $\triangle KAB$, we get PS || and = 0.5 AB = 0.5DC = DN. That means DNSP is a ||gm. Similarly, BMTP is also a ||gm.
Hence, $\angle BRK= \angle KTM = 90^0$. Similarly, $\angle DUK = \angle KSN = 90^0$.
Result follows from the fact that RBDU is cyclic.