The triangle $ABC$ is right angld at $A$. A line through the midpoint $D$ of $BC$ meets $AB$ at $X$ and $AC$ at $Y$. The point $P$ is taken on this line so that $PD$ and $XY$ have the same midpoint $M$. The perpendicular from $P$ to $BC$ meets $BC$ at T.
Prove that $AM$ bisects $\angle TAD$.
I have puzzled over this problem from my book on innovative Euclidean Geometry for months.
The book doesn't have solutions, only hints so you can imagine how frustrating this can be.
I would REALLY appreciate this if someone could solve it or at least make headway on it.
If you would like the hint provided by my book just ask. Thanks.