$Problem$ $2$.
$ABC$ is a triangle where $∠BAC = 90◦$. A line through the midpoint $D$ of $BC$ meets $AB$ at $X$ and $AC$ at $Y$, where $X$ and $Y$ are not on the same side of $BC$. The point $P$ is taken on $XY$ such that $PD$ and $XY$ have the same midpoint $M$. The perpendicular from $P$ to $BC$ meets $BC$ at $T$.
Prove that $AM$ bisects $∠TAD$.
National Math Camp Geometry Exam Problem-2
- Anindya Biswas
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Re: National Math Camp Geometry Exam Problem-2
$\textbf{Solution :}$
$D$ is the circumcenter of $BAC$.
"If people do not believe that mathematics is simple, it is only because they do not realize how complicated life is."
— John von Neumann
— John von Neumann