<PERSON>'s answer is correct, but I wanted to elaborate a bit further.
The Earth's axis is tilted about $23.5^\circ$ from being perpendicular to the plane of its orbit (as <PERSON> mentioned, 'obliquity' or 'axial tilt' is the name for this.) As a result of this tilt, at the solstices (the first day of winter or summer,) the sun is directly facing latitudes of either $23.5^\circ$ North (the first day of summer in the Northern hemisphere) or $23.5^\circ$ South (the first day of summer in the Southern hemisphere.) Areas between these two latitudes are known as the tropics, where the sun will be directly overhead at some point during the year.
For latitudes farther away from the equator than the tropics, while the sun will never be directly overhead, the first day of summer in a given hemisphere will be the day in which the incident angle of the sun is the most shallow (i.e. the sun is as close as it will come to being directly overhead) and the first day of winter is the day in which the incident angle will be the most steep (i.e. the sun will be the farthest from passing directly overhead.) For example, at a latitude of $40^\circ$, on the first day of summer, the sun will be only $16.5^\circ$ away from being directly overhead, while, on the first day of winter, it will be $63.5^\circ$ away from being directly overhead. Solar power density for areas directly under the sun is about $1.4\ \mathrm{kW/m^2}$, so this gives us:
Solar power density at noon on the first day of summer at 40 degrees latitude:
$$1.4\ \mathrm{kW/m^2} * \cos(16.5^\circ) = 1.4\ \mathrm{kW/m^2} * 0.959 = 1.34\ \mathrm{kW/m^2}$$
And for winter:
$$1.4\ \mathrm{kW/m^2} * \cos(63.5^\circ) = 1.4\ \mathrm{kW/m^2} * 0.446 = 0.62\ \mathrm{kW/m^2}$$
Additionally, the sun will spend much more time above the horizon on the first day of summer than on the first day of winter, again, due to the axial tilt of Earth. 15
The combination of these two effects results in the dramatic difference in average solar heating observed between summer and winter in latitudes that are distant from the equator. This local difference in solar heating through the year is the dominant factor in determining seasons as the Earth orbits the sun, not the comparatively minor differences in actual distance between the Earth and the sun.
For a graphical comparison, take a look at the solar power density measurement graphs on these pages, from a weather station located about $36.2^\circ$ North:
December 20, 2014 (a day before the solstice because the actual solstice was cloudy)
June 22, 2014 (again, a day off the solstice, due to clouds on the solstice.)
I would copy the graphs into this post, but they're copyrighted by that site, so you'll have to visit the links and scroll down to the power density graphs. Note that apparent noise in the graphs is primarily due to clouds briefly passing over.
The areas under these curves (i.e. total solar energy per unit area per day) are roughly $2.4\ \mathrm{kW\ hr/m^2}$ for the beginning of winter and $7.3\ \mathrm{kW\ hr/m^2}$ for the beginning of summer. At least, that's about what they would have been had there been no clouds in the way. 108