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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.

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Oh man! I posted the figure thinking it would be useful.;) – Inceptio Apr 13 '13 at 8:26
don't worry, (thank goodness the book had a picture in it or I would have got nowhere). :-) – John Marty Apr 13 '13 at 8:29
We would help you, anyway.:) – Inceptio Apr 13 '13 at 8:31
Using coordinates, I find that $\angle TAM = \angle DAM = \angle DXA - \angle DBA$. Now to find an elegant proof of this fact ... – Blue Apr 13 '13 at 9:22
Do you mean $DBA-DXA$, instead of the other way around? It looks a lot like, in general, $DBA>DXA$, namely. – HSN Apr 13 '13 at 9:32
up vote 6 down vote accepted

\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 $\angle TAM = \angle MAD$.


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Very Beautiful; thank you SO much. It's amazing how many clever little deductions you have to put together to reach such a result. Thanks again!!! – John Marty Apr 13 '13 at 10:53

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