1966 AHSME Problems/Problem 32

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Problem

Let $M$ be the midpoint of side $AB$ of triangle $ABC$. Let $P$ be a point on $AB$ between $A$ and $M$, and let $MD$ be drawn parallel to $PC$ and intersecting $BC$ at $D$. If the ratio of the area of triangle $BPD$ to that of triangle $ABC$ is denoted by $r$, then

$\text{(A) } \frac{1}{2}<r<1 \text{,  depending upon the position of P} \\ \text{(B) } r=\frac{1}{2} \text{,  independent of the position of P} \\ \text{(C) } \frac{1}{2} \le r <1 \text{,  depending upon the position of P} \\ \text{(D) } \frac{1}{3}<r<\frac{2}{3} \text{,  depending upon the position of P}\\ \text{(E) } r=\frac{1}{3} \text{,  independent of the position of P}$

Solution

$\fbox{B}$

See also

1966 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 31
Followed by
Problem 33
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