构造：通过一点作直线的垂线

The critical part of this construction is that in an isosceles triangle, the vertex (intersection of the two equal sides) lies on the perpendicular bisector of the opposite side.

From a point $P$ to a line $\ell$, draw a circle around $P$ with your compass that intersects $\ell$ at $Q$ and $R$. Since $PQ=PR$, $P$ lies on the perpendicular bisector of $QR$. Therefore, just construct $\ell'$ to be the perpendicular bisector of $QR$, and that is the unique perpendicular line to $\ell$ from $P$.

Note: to construct the perpendicular bisector, draw two circles of equal radius around $Q$ and $R$, intersecting at $X$ and $Y$. $XY$ is the perpendicular bisector.