Super hard Euclidean Geometry 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. 

 A: \begin{align*}
\angle TDM &= \angle YDC
\\ &= \angle DYA - \angle DCY \text{ (exterior angle = sum of opposite interior angles in triangle }DCY)
\\ &= \angle MAY - \angle DCY \text{ ($M$ is centre of circle through $XAY$, so $\angle MYA = \angle MAY$)}
\\ &= \angle MAY - \angle DAY \text{ ($D$ is centre of circle through $BAC$, so $\angle DCY = \angle DAY$)}
\\ &= \angle MAD
 \end{align*}
But $\angle TDM = \angle MTD$ (because $M$ is centre of circle through $PTD$, so $MD = MT$). Thus $\angle MTD = \angle MAD$, and so $MTAD$ is a cyclic quadrilateral. And $MD = MT$. Hence $\angle TAM = \angle MAD$.
QED
A: Alright, we have to prove that angle TAM equals angle MAD. Because XAY and BAC are right triangles,


*

*DAY=DCA=alpha; ABD=BAD=90-gamma

*MAY=MYA=gamma+alpha ; MAX=MXA=90-gamma-alpha
As you may have noticed, I´m calling angle MAD alpha. Then, with these properties in hand, we have:


*

*ADB=2gamma (external angle)

*YDC=alpha (We know that angle DYA equals gamma+alpha, and it`s external angle of triangle DYC. Since angle DCY equals gamma, angle YDC can only be alpha)

*TDX=alpha (Opposite to angle YDC)

*ADY=180-2gamma-alpha (sum of angles in triangle ADY)

*ADT=2gamma (shallow angle)


But we have that since M id the midpoint of segment PD, PM=MD=MT. Now, we just have to prove that TMDA is a cyclic quadrilateral, since angle TDM equals alpha equals angle MTD (because TM=MD), which gives us that angle TMX=2alpha. Please note that I´m saying that we have to prove that TMDA is a cyclic quadrilateral because it would gives us the fact that angle TAM equals alpha, since that sum of opposite angles in a cyclic quadrilateral equals 180 degrees.
But note that we´ve already accomplished that: by the sums of angles in triangle MAY, we have that angle PMA equals 2gamma+2alpha. But angle PMT equals 2alpha. Then, angle TMA equals 2gamma. Hey, angle TDA equals 2gamma, too! Also, angle MTD equals angle MAD equals alpha.
To conclude, because TMDA is a cyclic quadrilateral, angle TMP equals angle TAD. So, angle TAM equals alpha, and it`s done.
