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Topics in Euclidean Geometry

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Topics in Euclidean Geometry Topics in Euclidean geometry Congruence - suggested problems - solutions A pair of proofs. Anything out of Euclid’s Book I is allowable for justification, and if you need to modernize things a bit by adding variables and doing algebra, that’s fine too. ∼ P1: Given 4DEF equilateral, 6 EDA = 6 BEF , EA = DC, BF = DC, prove 4ABC is equilateral, and that CF = AD. This one’s involved (which is why it’s here), and I’m going to do something halfway between a narrative and a column proof so I can explain the logic as I go. The key is that it still follows the essential “triangle proof logic” - work with what you HAVE before worrying about what you want, see what triangles you can get congruent, and use the congruent triangles to claim the pieces you need at the end. The first given is that the big triangle is equilateral. What’s that buy you? Three equal sides, and three equal angles. Since 4DEF equilateral, we have DE = EF = F D. In addi- tion, since equilateral triangles are also equiangular (Proposition 8), m6 EF D = m6 F DE = m6 DEF . The second given focuses your attention on the highlighted triangles. At this point, you have a side and an angle in each. ∼ 6 EDA = 6 BEF The third and fourth given can be combined (and yes, just do that; you don’t have to justify obvious algebra, such “two things equal to the same thing are equal to each other”, which is what this is (transitivity)) EA = DC = BF Now, you look at that, and think, well that’s no good. The thing you want to notice is that with the highlighted triangle, you don’t have side-angle-side. The angle isn’t between the known sides (i.e., you’ve got DC = EA, but you need DA = AB - the short legs of the triangles are the other side of the angle. So you go back and say hmm, I know the “big” angles are all equal (from equilateral triangle). Each of those angles is split, ∼ and you know that 6 EDA = 6 BEF . So it must also be the ∼ case that 6 FDA = 6 DEB. This puts you in a different pair of triangles. The justification for the angle shift is just showing a bit of algebra: Since m6 F DE = m6 DEF and m6 EDA = m6 BEF , and since m6 F DE = m6 FDA + m6 EDA and m6 DEF = m6 BEF + m6 DEB, we have m6 FDA = m6 DEB, or ∼ 6 FDA = 6 DEB. (This is obviously easier to see immediately in the diagram, and a pain to keep the letters straight writing it all out. I think I didn’t typo.) NOW, you can claim that those two triangles are congruent. ∼ By the SAS = SAS proposition, 4FDA = 4DEB. This doesn’t seem to be getting us an equilateral triangle yet, but it’s a step in the right direction - now that we’ve got two trian- gles congruent, we can claim all their parts. The fact that we’re going for the equilateral triangle in the center suggests we should probably figure out how to bring the third triangle into the picture. Originally, we had the two marked angles congruent, but we’ve also got the two triangles congruent. This lets us bring a third angle into the picture. ∼ ∼ Since 4FDA = 4DEB, 6 EDA = 6 DFC (and also to 6 F EB). And THAT lets us move into the third triangle (by subtraction from the “big” angle again). ∼ ∼ And, since 6 EDF = 6 DF E, 6 FDA = 6 EFC (and also to 6 DEB). (You can skip the algebra, it’s redundant, and pretty obvious where it’s coming from.) The third triangle comes in congruent to the other two by angle-side-angle. ∼ ∼ By the ASA = ASA proposition, 4EFC = 4FDA = 4DEB You already were given that these sides were equal. Having the triangles congruent tells you these sides are equal (the short leg of each triangle). ∼ ∼ Since 4EFC = 4FDA = 4DEB, FC = DA = EB. (Which completes part of the proof - you have CF = AD here.) And by subtraction, these sides must be equal. Since F B = DC = EA and FC = DA = EB, and since F B = FC + CB, DC = DA + AC, and EA = EB + BA, CD = AC = BA. Therefore, 4ABC is equilateral. And there’s the last of it! P2: Given BF bisects 6 CBA, CD bisects 6 ACB, m6 CBA = m6 ACB, prove BF = CD. This is a “split the triangles to see better” proof. It may also make things a bit easier to in- troduce a little algebra. Since you’re working on getting BF = CD, you need to be getting ∼ 4BCD = 4CBF . Start with the first given. What does it tell you? ∼ Given BF bisects 6 CBA, we have 6 CBF = 6 ABF . We also have 2m6 CBF = m6 CBA. (Get all the info you can out of the bisector. The reason I focus on the measure of 6 CBF is because it’s in one of the triangles we’re going for.) Next given, same thing. ∼ Given CD bisects 6 ACB, we have 6 BCD = 6 ACD. We also have 2m6 BCD = m6 BCA. This isn’t getting you anything yet with respect to the separated triangles, but push on. Because the next given gets you two things. First, some angles pair up immediately, but also, since you’ve just discovered that you’ve cut each of two equal angles in half, the halves must be equal as well. That means the previously unrelated angles in the separated triangles are in fact congruent. Given m6 CBA = m6 ACB, by substituting, we have 2m6 CBF = 2m6 BCD, and therefore m6 CBF = m6 BCD. Which pretty much does it. The triangles share a base, so by ASA they’re congruent, and therefore so are their remaining matching parts. (You may recall something from doing two column proofs - “CPTCE” - “corresponding parts of congruent triangles are equal”. I wouldn’t use it in a narrative proof, generally. It’s part of the idea of congruence.) ∼ Since BC = CB, we have 4BCD = 4CBF by ASA = ASA, and therefore CD = BF.
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