# Thread: Euclid's Proposition 35

1. ## Euclid's Proposition 35

Parallelograms which are on the same base and in the same parallels equal one another.

Is this proposition true if the parallelograms in question is a square?

Let ABCD be a square, and from base cd, let another parallelograms be created. Let's call the new parallelograms EBCF. Since it's a parallelograms, EF is equal to BC.

According to Euclid, the square ABCD is equal to EBCF. And here is my problem: how can they be equal? Unless drawn on the same side, EBCF will have two diagonals which will be more than the bases. EBCF will not be a square, how can it be equal to ABCD?

2. "Let ABCD be a square, and from base cd, let another parallelograms be created. Let's call the new parallelograms EBCF. Since it's a parallelograms, EF is equal to BC."

Something wrong there.
If If the two paralleograms have the same base, then the base CD of the square ABCD must be the same base of parallelogram EBCF. But since EBCF does not even have CD, how can the two parallelogram be on the same base?
The only "same" side on ABCD and EBCF is BC. I thought you said "...from the base cd..."

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"Parallelograms which are on the same base and in the same parallels equal one another."

What that means is that if two parallelograms have the same base, and their other bases are in-line or aligned or in the same line, (or the perpendicular distance from the same base to their other bases is the same, or the two parallelograms have the same perpendicular height or altitude), then the areas of the two parallelograms are equal.

That is true.
Because whatever is the shape (or sway, or leaning) of a parallelogram, its area is always (base)*(altitude).

3. Originally Posted by ticbol
Something wrong there.
If If the two paralleograms have the same base, then the base CD of the square ABCD must be the same base of parallelogram EBCF. But since EBCF does not even have CD, how can the two parallelogram be on the same base?
The only "same" side on ABCD and EBCF is BC. I thought you said "...from the base cd..."
Sorry, my mistake. (Apologies again, I can't insert the image I drew. Just picture AD is in line with EF.)

According the Euclid's Proposition 35, parallelogram ABCD is equal
to parallelogram EBCF.

Parallelogram ABCD was drawn to be a square, therefore, AB = BC = CD = DA.
For EBCF to be equal to ABCD, all sides of EBCF must be equal, because the area
of a square is s*s.

Is EB = BC = CF = FE?

EF = BC because EBCF is a parallelogram and EF & BC are opposite sides.
Therefore, EF = AB = AD.

The triangle EAB has the sides AB, EA and BC, and it is a right triangle, since
angle EAB is an angle of a square. The triangle in my figure has the line AB = 1,
line EA > 1, since AD, which equals one, is a part of EA. Using the pythagorean
theorem, the hypotenuse EB is longer than either EA or AB.

EB is therefore longer than EF or BC. Therefore, ABCD != EBCF.

What is wrong with my line of reasoning? Years of C++ Programming has taught me
that before I assume that there's somethin wrong with the OS, I should first
check my codes. All the more with Euclid, the error is certainly in my reasoning,
but I cannot figure where I went wrong.

4. "...Just picture AD is in line with EF."

So, the same base is BC, allright, and the other bases are AD for the square and EF for the parallelogram.

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"For EBCF to be equal to ABCD, all sides of EBCF must be equal, because the area of a square is s*s."

1) If EBCF is a square also, then all sides of EBCF are equal, and ABCD and EBCF are both squares, one on top of the other, and only one square can be seen on the image/drawing/figure.

2) If EBCF is not a square, then all its sides are not equal if EF is in line with AD. For all the sides of non-square EBCF to be equal, EBCF must be a rhombus, but then EF will not be in line with AD---here, EF will be closer to BC than AD. (Imagine what will happen if you sway/lean the now square ABCD so that ABCD won't be square anymore. AD will come closer to BC.)

3) Then, if EBCF were a rhombus, its area won't be s*s. Its area will be less than s*s.

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"Is EB = BC = CF = FE?"

No.
BC = EF, and EB = CF, because EBCF is a parallelogram, and not a square nor a rhombus.

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"EF = BC because EBCF is a parallelogram and EF & BC are opposite sides.
Therefore, EF = AB = AD."

The first sentence is correct.
The last one is wrong. Although AB = AD = BC in the square ABCD, the "AB = AD = BC = EF" is wrong because AB has no relation to the parallelogram EBCF.
AB = AD = BC only on the square.

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"The triangle EAB has the sides AB, EA and BC, and it is a right triangle, since
angle EAB is an angle of a square. The triangle in my figure has the line AB = 1, line EA > 1, since AD, which equals one, is a part of EA. Using the Pythagorean theorem, the hypotenuse EB is longer than either EA or AB."

Whoa. Here you really knocked me down. I cannot complete/finish explaining this paragraph.

How could triangle EAB contain BC? BC is a side of triangle EAB?

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"What is wrong with my line of reasoning?"

Ummm,...

5. Here's the figure... from the figure, we see that ABCD is a square, and EBCF, a parallelogram. They can't be equal in area, but Euclid Proposition 35 states that they are equal since they have both the same base, BC, and in the same parallel, AF || BC. Again, where did I go wrong here?

6. The first sentence is correct.
The last one is wrong. Although AB = AD = BC in the square ABCD, the "AB = AD = BC = EF" is wrong because AB has no relation to the parallelogram EBCF.
AB = AD = BC only on the square.
The relation between ABCD and EBCF is the base BC. Since AB = BC and BC = EF, then EF = AB. Things which are equal to the same thing are equal.

7. How could triangle EAB contain BC? BC is a side of triangle EAB?
Another apology, I'm really sorry for my clumsiness. What I meant to say is that triangle EAB contains the side EA, AB and BE. Please refer to the figure.

8. Umm, now, with that that figure, I see what you mean.

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But first, your
"EF = BC because EBCF is a parallelogram and EF & BC are opposite sides.
Therefore, EF = AB = AD."
is correct. I got lost somewhere. I thought one of the sides mentioned there is a slanting side of EBCF.

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So in the figure you are required to prove the areas of ABCD and EBCF are equal.

Here is one way. All statements only. No reasons (You can provide the reasons).

Let us call the intersection of EB and CD as point G.

1. Triangle BCG = Triangle BCG

If we can prove ABGD = EGCF, then it follows that ABCD = EBCF.

2. DE = DE
3. AD = EF
4. AD +DE = DE +EF
So,
5. AE = DF

We are trying to prove here that triangle ABE = triangle DCF,
6. AB = CD
So,
7. triangle ABE = triangle DCF
(Area of triangle = base*altitude. Triangles ABE and DCF have equal bases and altitudes.)
Then,
8. Triangle DGE = triangle DGE
So,
9. (triangle ABE) -(triangle DGE) = (triangle DCF) -(triangle DGE)
So,
10. ABGD = EGCF
Therefore,
11. ABCD = EBCF -------------proven.

That is because ABCD = ABGD + BCG.
And EBCF = BCG +EGCF.

9. The height of the parallelogram is also one. The diagonal part is not the height rather... see the figure.