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.
\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