Let $\triangle ABC$ have $M$ as the midpoint of $BC$ and let $P$ and $Q$ be the feet of the altitudes from $M$ to $AB$ and $AC$ respectively. Find $\angle BAC$ if $[MPQ]=\frac{1}{4}[ABC]$ and $P$ and $Q$ lie on the segments $AB$ and $AC$.
1. Let $\triangle ABC$ be given with $M$ as the midpoint of $BC$. Let $P$ and $Q$ be the feet of the perpendiculars from $M$ to $AB$ and $AC$ respectively.
2. We are given that the area of $\triangle MPQ$ is $\frac{1}{4}$ of the area of $\triangle ABC$, i.e., $[MPQ] = \frac{1}{4}[ABC]$.
3. Since $M$ is the midpoint of $BC$, $M$ divides $\triangle ABC$ into two smaller triangles $\triangle ABM$ and $\triangle ACM$ of equal area. Therefore, $[ABM] = [ACM] = \frac{1}{2}[ABC]$.
4. Consider the altitudes from $M$ to $AB$ and $AC$. Since $P$ and $Q$ are the feet of these altitudes, $MP \perp AB$ and $MQ \perp AC$.
5. The area of $\triangle MPQ$ can be expressed as:
   \[
   [MPQ] = \frac{1}{2} \times MP \times MQ \times \sin(\angle PMQ)
   \]
6. Since $M$ is the midpoint of $BC$, $MP$ and $MQ$ are the altitudes of $\triangle ABM$ and $\triangle ACM$ respectively. Therefore, $MP$ and $MQ$ are half the lengths of the altitudes from $A$ to $BC$.
7. Let $h$ be the altitude from $A$ to $BC$. Then $MP = MQ = \frac{h}{2}$.
8. Substituting $MP$ and $MQ$ into the area formula for $\triangle MPQ$, we get:
   \[
   [MPQ] = \frac{1}{2} \times \frac{h}{2} \times \frac{h}{2} \times \sin(\angle PMQ) = \frac{h^2}{8} \sin(\angle PMQ)
   \]
9. We are given that $[MPQ] = \frac{1}{4}[ABC]$. The area of $\triangle ABC$ is:
   \[
   [ABC] = \frac{1}{2} \times BC \times h
   \]
10. Therefore, we have:
    \[
    \frac{h^2}{8} \sin(\angle PMQ) = \frac{1}{4} \times \frac{1}{2} \times BC \times h
    \]
11. Simplifying, we get:
    \[
    \frac{h^2}{8} \sin(\angle PMQ) = \frac{BC \times h}{8}
    \]
12. Dividing both sides by $\frac{h}{8}$, we get:
    \[
    h \sin(\angle PMQ) = BC
    \]
13. Since $M$ is the midpoint of $BC$, $BC = 2 \times BM$. Therefore:
    \[
    h \sin(\angle PMQ) = 2 \times BM
    \]
14. Since $BM = \frac{BC}{2}$, we have:
    \[
    h \sin(\angle PMQ) = BC
    \]
15. Therefore, $\sin(\angle PMQ) = 1$, which implies that $\angle PMQ = 90^\circ$.
16. Since $\angle PMQ = 90^\circ$, $\angle BAC$ must be $90^\circ$.

The final answer is $\boxed{90^\circ}$.