A triangle is half a parallelogram

0:00Every triangle has a base and a height, and almost everyone meets those words in the wrong order. The height is not a property of the triangle. It belongs to a base that somebody chose. Pick a side, call it the base, and the height is the perpendicular distance from it up to the far corner. Choose a different side and you get a different height. A triangle has three of them, and all three give the same answer.

0:30Over six thousand one hundred and seventy-six triangles, every one of the three bases returns the same number, and that number is the area. A height is also never one of the sides by accident. It is at most as long as the two sides that meet at its own corner, and it equals one of them only when that corner is a right angle. Start with the easiest triangle there is: one with a right angle in it.

1:02Stand it inside the rectangle that fits snugly around it. The diagonal cuts that rectangle into two pieces, and the two pieces are the same triangle twice over. So the triangle is exactly half the rectangle, which is half of base times height. That argument is airtight. It is airtight about one shape, and most triangles are not it. Now a triangle with no right angle anywhere in it. Drop a perpendicular from the top corner straight down to the base.

1:38That single cut splits the triangle into two right-angled pieces, and it splits the base into two parts. Box each piece in its own rectangle, and each piece is half of its own box. Add the two halves: half the first part times the height, plus half the second part times the height. The two parts add up to the whole base, so the total is half of base times height again.

2:06The two pieces do not overlap, and their areas add up to the whole triangle. There is a hole in that picture, and it is not a small one. The argument needs the foot of the perpendicular to land on the base, and sometimes it does not. Lean the top corner far enough sideways and the perpendicular comes down outside the base altogether. Then there is no triangle inside the figure to cut off.

2:36Ask every triangle in the sweep where its three feet land, and there are exactly three answers. If every corner is sharp, all three feet land inside. If there is a right angle, one lands inside and two land exactly on a corner. If one corner is blunt, one lands inside and the other two land off the base entirely. So what do you do? One answer is to subtract.

3:06Extend the base out to the foot, take the bigger triangle the argument does work for, and take away the extra piece hanging outside. It comes out exactly right, every time. But look again at those three patterns. Whatever the triangle, one base always works, and you can say in advance which. Put the base on the longest side. Over all six thousand one hundred and seventy-six triangles, the foot lands on the longest side every single time.

3:40The broken picture was never a problem with the argument. It was a problem with the side somebody chose. There is a second proof that needs no choosing. Take the triangle, make a copy of it, and turn the copy through half a turn about the midpoint of one side. The two of them close into a parallelogram, with no gap and no overlap. A half turn moves nothing but position, so every distance survives it.

4:13And a half turn sends every line to a line parallel to it. That is why the sides pair off: each side of the copy comes out parallel to the side it was made from. The parallelogram stands on the same base with the same height, and the triangle is half of it. Half of base times height, proved a second time, with no condition on where anything lands. Now the result worth carrying away.

4:44Take any triangle, mark the middle of one side, and join that midpoint to the opposite corner. That line is called a median, and it cuts the triangle into two. The two pieces have equal area, always. The reason takes one sentence. The two pieces have equal bases, because the midpoint halves the side, and they share a height, because the far corner is the same corner for both. Equal base, equal height, equal area.

5:16Checked on every triangle in the sweep, on all three of its medians. That ought to feel strange, and here is why. The two pieces are not the same shape. Take a base of eight with the cut at four, and put the apex six units up but only one across, near one end. Both halves have area twelve, exactly twelve. But one has a long side of the square root of thirty-seven and the other of the square root of eighty-five.

5:49They are not congruent, and not even scaled copies of one another. What they share is the only two things area can see: base four, and height six. Across the sweep, the triangles whose median halves really are the same shape are exactly the ones already symmetric about that median, two hundred and fifty-four out of six thousand one hundred and seventy-six. Fewer than one in twenty. Area is blind to shape, and the median is where that becomes impossible to ignore.

6:26So equal area does not mean the same shape. What it does mean is more than you would guess. Any two straight-sided figures of equal area can be cut into finitely many pieces that rebuild one another. That is a real theorem, and not an obvious one. Here is the simplest instance, performed rather than promised. Cut a triangle along the line half way up, then cut straight down the height, and turn the two top pieces down about the midpoints of the two sides.

7:01Three pieces, no overlap, and they fill a rectangle on the same base of half the height. The matching statement about solids is false: two solids of equal volume cannot always be cut into each other. Flat shapes are the lucky case. Once area sees only a base and a height, a whole class of questions collapses. Take a rectangle ten across and eight tall, put a triangle on the left side, and let its top corner sit anywhere on the right side.

7:37The area is forty wherever you put it. Seventeen positions tried, one answer. Take a point anywhere along a median and join it to the two ends of the base: the two triangles it makes are equal, everywhere along that median. Take a point anywhere inside a square and join it to all four corners. The two triangles on one pair of opposite sides have the same total area as the other two.

8:07One to one, at every point tried. That one holds in any parallelogram, and it fails the moment the sides stop pairing off. Here is one more, with a warning attached. Halve one side of a triangle, put a point along a second side, and slide the third corner along the first until the join comes out parallel. The triangle left over is half the original, and that part is true at every position.

8:39But the construction asks you to slide a corner, and the corner runs off the end of the side unless your point is at least half way along. Of twelve positions tried, seven leave it on the side and five send it past the end. The tidy drawing shows one of the seven. It is the same lesson as before: a picture that works is not yet an argument that works.

9:07Two last things. Double every side of a triangle and the area does not double. It goes up four times. Treble the sides and it goes up nine. And the copies genuinely fit, which is the sharper half of the question. Join the midpoints of the doubled triangle and it falls into four pieces, every one congruent to the original, none overlapping. Trisect the sides instead and you get nine.

9:38Finally, a question that looks like all the others and is not. Among the rectangles with perimeter forty, which has the largest area? The ten by ten square, with area one hundred. Which has the smallest? None of them does. The area can be pushed below any target you name and never reaches the bottom, so the question has no answer, and noticing that is worth more than a number.