Really really hard and old Euclidean geometry problem

Question

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

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

Weird thing about this problem: It was asked in ahsme 1966 and on the AOPS site for ahsme 1966, they've uploaded solutions for every single one of the 40 problems, except this one, they've just stated that the correct answer is $B$

Anyways, to solve this, I tried using area ratios of similar triangles(the triangles with the bases as the two parallel lines) but that didn't lead to anything. I tried writing down all the ratios and converting ratios of areas to ratios of bases, but all that did was create a messy system of equations with way more unknowns than equations. Please help me with this one

Answer 1

First Approach

Because $PC\parallel MD$, we have by similar triangles that $$ \frac{BM}{BD}=\frac{BP}{BC}\tag1 $$ We can compute the areas with $$ \left|\triangle BPD\right|=\frac12|BP|\,|BD|\sin(B)\tag2 $$ and $$ \begin{align} \left|\triangle BMC\right| &=\frac12|BM|\,|BC|\sin(B)\tag3\\ &=\frac14|BA|\,|BC|\sin(B)\tag4\\ &=\frac12\left|\triangle ABC\right|\tag5 \end{align} $$ However, by $(1)$, $$ \overbrace{\frac12|BP|\,|BD|\sin(B)}^{\left|\triangle BPD\right|}=\overbrace{\frac12|BM|\,|BC|\sin(B)}^{\left|\triangle BMC\right|}\tag6 $$ Thus, $$ \left|\triangle BPD\right|=\frac12\left|\triangle ABC\right|\tag7 $$

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Second Approach (from achille hui's comment)

Because both $\triangle MDP$ and $\triangle MDC$ have base $MD$ and the same altitude, i.e. the distance between $MD$ and $PC$, $$ \left|\triangle MDP\right|=\left|\triangle MDC\right|\tag8 $$ Thus, including $\left|\triangle BMD\right|$, we get $$ \left|\triangle BPD\right|=\left|\triangle BMC\right|\tag9 $$ Furthermore, because $|BM|=\frac12|AB|$, $$ \left|\triangle BMC\right|=\frac12\left|\triangle ABC\right|\tag{10} $$ Combining $(9)$ and $(10)$, we get $$ \left|\triangle BPD\right|=\frac12\left|\triangle ABC\right|\tag{11} $$

Answer 2

This is kind of a wise guy solution, but I can't resist posting it. Without loss of generality $$A=\begin{pmatrix}-1\\0\end{pmatrix},B=\begin{pmatrix}1\\0\end{pmatrix},M=\begin{pmatrix}0\\0\end{pmatrix}$$ and $$C=\begin{pmatrix}x_0\\y_0\end{pmatrix},\ y_0\neq0$$ Multiply all points by any non-singular transformation $T$. $T$ take parallel lines to parallel lines and $TM=\frac12(TA+TB)$ so that we get a diagram of the same type. Since $T$ multiplies areas by $|\det T|$, the ratio of the two triangles is the same in all figures, so only B and E are possible answers. (Of course, I really should say that we can get from any valid diagram to any other, but this is clear.)

In the limiting case where $P$ coincides with $A$, the ratio is $\frac12$ by similar triangles, so B is the answer, by continuity.