For this first proof (I.34), we draw two pairs of parallel line segments (\(\small\overline{\mathtt{AC}} \parallel \overline{\mathtt{BD}}\) and \(\small\overline{\mathtt{AB}} \parallel \overline{\mathtt{CD}}\)) and a diagonal, which Euclid calls a diameter (\(\small\overline{\mathtt{BC}}\)).
What we want to show in this proof is that "in parallelogrammic areas the opposite sides and angles are equal to one another, and the diameter [diagonal] bisects the areas." (Hover and click on the chalkboard if you want.)
Again, since \(\small\mathtt{AC}\) is parallel to \(\small\mathtt{BD}\), and \(\small\mathtt{BC}\) has fallen upon them, the alternate angles \(\small\mathtt{ACB}\), \(\small\mathtt{CBD}\) are equal to one another [I.29 again].
Therefore \(\small\mathtt{ABC}\), \(\small\mathtt{DCB}\) are two triangles having the two angles \(\small\mathtt{ABC}\), \(\small\mathtt{BCA}\) equal to the two angles \(\small\mathtt{DCB}\), \(\small\mathtt{CBD}\) respectively, and one side equal to one side, namely that adjoining the equal angles and common to both of them, \(\small\mathtt{BC}\); therefore they will also have the remaining sides equal to the remaining sides respectively, and the remaining angle to the remaining angle.
That final "therefore" from Euclid in the quotation comes to us courtesy of the Angle-Side-Angle Congruence Theorem (I.26). So, we know that \(\small\Delta\mathtt{ABC} \cong \Delta\mathtt{DCB}\), and the corresponding congruent sides in each triangle are those opposite the corresponding congruent angles.
The diagram really tells the rest of the story. The opposite angles are equal to each other, and the diagonal, or "diameter," divides the "parallelogrammic area" into two equal triangular areas, quod erat demonstrandum.
Same Base, Same Parallels
Next up is I.37, where we want to show that "triangles which are on the same base and in the same parallels are equal [in area] to one another."
First, though, we can do a demonstration. Click Run below to start the simulation at the left, and click Reset to stop it and start over.
\(\small\Delta\mathtt{ABC}\): Area \(\small\Delta\mathtt{DBC}\): Area
The way this works is that the computer chooses a random point on the chalkboard. For each triangle, it records whether that point falls inside or outside and then displays the fraction of points that fall inside the triangle (and of course shows you all the points that land inside the triangles [but with repeats, so this test has some flaws besides being probabilistic]). I know that the width of \(\small\Delta\mathtt{DBC}\) is 190 pixels, and its height is 170 pixels. The chalkboard measures 300 × 300 square pixels. You can tell from that whether the numbers given in your run of the simulation are close to correct.
That was nice, but let's prove that the two triangles with the same base and between the same parallel lines must be congruent.
I'll partly follow Euclid's lead in re-sketching the diagram, since he got to this first—about 2300 years ago:
Euclid gets this conclusion from I.35, which we haven't discussed here. But the proof you saw in the first section (I.34) gets us the next step. That is, \(\small\Delta\mathtt{ABC}\) is half the area of parallelogram \(\small\mathtt{EBCA}\) because \(\small\overline{\mathtt{AB}}\) bisects it, and \(\small\Delta\mathtt{DBC}\) is half the area of parallelogram \(\small \mathtt{DBCF}\) because \(\small\overline{\mathtt{DC}}\) bisects it. This means that \(\small\Delta\mathtt{ABC}\) is equal in area to \(\small\Delta\mathtt{DBC}\), since "halves of equal things are equal to one another."
And Finally, Ladies and Germs, I.41
We can now use our conclusions in I.34 and I.37 above to prove I.41: that "if a parallelogram have the same base with a triangle and be in the same parallels, the parallelogram is double of the triangle."
Euclid starts with parallelogram \(\small\mathtt{ABCD}\) which has the same base as \(\small\Delta\mathtt{EBC}\), and both are "in the same parallels," \(\small \overleftrightarrow{\mathtt{AE}} \parallel \overleftrightarrow{\mathtt{BC}}\).
The diagonal segment \(\small\overline{\mathtt{AC}}\), along with I.37 above, tells us that \(\small\Delta\mathtt{ABC}\) and \(\small\Delta\mathtt{EBC}\) have the same area, because they are in the same parallels and have the same base. And I.34 tells us that \(\small\Delta\mathtt{ABC}\) is half the area of parallelogram \(\small\mathtt{ABCD}\). Therefore, \(\small\Delta\mathtt{EBC}\) is also half the area of parallelogram \(\small\mathtt{ABCD}\).
Image-mask credit: John Kinsella
