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, 2012

Geometry Difficulty 4.3 AIME Prove it India

Let $ABCD$ be a trapezium with $AB \parallel CD$ . Let $P$ be a point on $AC$ such that $C$ is between $A$ and $P$ ; and let $X$ , $Y$ be the mid-points of $AB$ , $CD$ respectively. Let $PX$ intersect $BC$ in $N$ and $PY$ intersect $AD$ in $M$ . Prove that $MN \parallel AB$ .

Solution

Observe that
$\frac{BN}{NC} = \frac{[PNB]}{[PNC]}$
However $[PNB] + [XNC] = [PXB] = [PXA] = [PCN] + [ACN] + [AXN]$ . Since $[XNB] = [AXN]$ , we obtain $[PNB] = [PNC] + [ACN]$ . Thus
$\frac{BN}{NC} = \frac{[PNC] + [ACN]}{[PNC]} = 1 + \frac{[ACN]}{[PNC]} = 1 + \frac{AC}{PC}.$
Similarly, we can prove that
$\frac{AM}{MD} = 1 + \frac{AC}{CP}.$
Comparison shows that $AM/MD = BN/NC$ . We conclude that $MN \parallel AB$ .

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