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**The Struggling Mathematician’s Guide to Solving Proofs**

By Mounir Lazzouni

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Introduction Thank you for choosing this presentation to help you refine your skills in doing proofs! This guide will teach you how to do generic proofs, how to get past mathematician’s block in doing proofs, and strategies to make proofs easier and even fun with the right attitude! This presentation is meant for those who are currently enrolled in a geometry class and need help with the challenging unit(s) of proofs. Those who are not in a geometry class may also benefit, although not as much and there are no guarantees.

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Getting Started Proofs can be one of the most difficult and time consuming parts of geometry, which has stained their reputation as the part of math everybody hates. That’s why it’s important to get a lot of practice with proofs so that they aren’t as hard or time consuming, and they can even be fun if you have the right attitude. One problem a lot of students face when solving a proof is lack of motivation. To deal with this, try to find something fun about the proof, because the main reason that people hate proofs is because they find them boring. If you like puzzles, you can think of proofs as puzzles since they take thought to solve and there are many different types of them. This helps a lot of people.

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**Getting Started (cont.)**

Another way to deal with this is to reward yourself whenever you solve a proof. I would award myself with a snack or candy every time I finished three proofs, which would be about every thirty minutes. You should do whatever you have to do in order to make you want to do proofs, or despise of them less. Even with the best motivation, one can still dread proofs. As a result, one may not practice proofs and remain mediocre at them, which adds to the tedium. The number one way to stop this is to practice proofs often. It may sound ridiculous, but if you keep yourself sharp on proofs you end up saving time, even after counting the time you spent doing proofs, because you spend so much less time when you have to do proofs in class or for homework.

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**Getting Started (cont.)**

Not only that, but you get the added bonus of when you need to do proofs on a final or test, you’re better at it, have a higher chance of getting it correct, and don’t need to worry about running out of time since you will be so much ahead the majority of people, which are those who don’t practice proofs. It’s almost a no brainer, you save time and get better grades, why would you not do it? Do you not have time? Whenever you have downtime in class or there’s a commercial break, get out the geometry book and pick a proof. It may even become fun and something that you look forward to doing. Either way, it’s all about your attitude towards doing the proofs, if you tell yourself that you want to do them and that you like doing proofs, it will make things a whole lot easier.

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How to do Proofs All of this talk about getting better at proofs by practicing them may make the reader wonder how they’re supposed to practice proofs if they don’t even know how to do them. Most people reading this PowerPoint have already taken a geometry class or are in the process of taking one, and probably have already learned a few proofs. If you do not have access to theorems, please visit

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How to do Proofs (cont.) The basic idea behind a geometric proof is to prove that two shapes have certain properties in relation to each other using theorems. A theorem is a theoretical proposition, statement, or formula embodying something to be proved from other propositions or formulas. If you are taking a geometry class, and that’s why you need help with theorems, then you most likely have a geometry textbook, if not at home then at school. Your textbook will have all of the theorems that you are expected to use on the proofs that you do for class. Again, if you don’t have access to theorems, visit

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**Example 1 Let’s start with an easy Algebraic proof.**

Supply the missing step in the following proof. Given: 4x – 6y = –8; x = 10 Prove: 8 = y A) –6y = 32; Multiplication Property of Equality B) –6y = –48; Subtraction Property of Equality C) –6y = –48; Multiplication Property of Equality D) –6y = 32; Subtraction Property of Equality To go about solving this problem, you would first substitute x for 10, which is already done for you. Next, you would then write the property that allows you to do that, the substitution property, which has also already been done. Next, combine like terms, subtract 40 from both sides to get 40 on the side of the other number not associated with a variable. In a multiple choice situation like this one, you can already eliminate 2 choices, a and d, because -6y = -48. To get to the next step, we must multiply both sides by -1/6 to get y by itself. We now have the next step on the other side of the blank. We had to multiply to get to that step, by using the Multiplication Property of Equality, which gets you the answer, c. Or in a multiple choice situation, you could eliminate b, since we did not need to subtract, or use the Subtraction Property of Equality.

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Example 2 3. Complete the proof. Given: , DA⊥AB, CE⊥BE. Prove: ΔADB ≅ ΔEBC. A) a. Distance Formula; b. given; c. HL Theorem B) a. Distance Formula; b. CPCTC; c. HL Theorem C) a. Distance Formula; b. given; c. SSS Theorem D) a. Distance Formula; b. CPCTC; c. ASA Theorem I explain this proof on the next slide.

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Example 2 (cont.) This proof is obviously much more difficult than the previous. If you find it too hard you may want to practice more of the algebraic proofs before going on to geometric proofs. Another thing that’s different about this proof besides being more difficult and being geometric is it’s a different style. This style is not required but it is recommended for geometric proofs because it’s simpler and easier to read for geometry, as long as using more paper is no problem. At any rate, to go about solving this proof, first you must gather as much information in addition to the givens as possible. We can see that by using the distance formula, it is possible to find that both CB and DB are equal to 5. The distance formula is , where x is the x coordinate of the equation with that number, and y is the y coordinate of the equation of that number. Since we used the distance formula, that is the answer to the first blank. Next we can use the transitive property to state that CB=5=DB, therefore CB=DB if you take out the 5. Because CB=DB, by using the definition of congruent segments, we can state that line CB ≅ line DB. We also are given that ∠ DAB and ∠ CEB are right angles. By the definition of a right triangle, we can also say that ΔDAB and ΔCEB are right triangles. We can finally conclude that the triangles are congruent by the Hypotenuse Leg (HL) Theorem, because we have the hypotenuse and leg.

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Example 3 Complete the proof. Given: bisects ∠URS and bisects ∠UTS. Prove: ΔURT ≅ ΔSRT. Explanation on next slide.

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Example 3 (cont.) Obviously, this proof is a step up from Example 2, although it is still of the intermediate level. Try to do this proof on your own, you can check your work on the next slide.

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Example 3 (cont.) Your proof should look like this:

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**Sources dictionary.com**

typeupsidedown.com

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Farewell Hopefully you enjoyed this and will apply what you’ve learned in this PowerPoint, or at least try it. If it’s had such an impact on you that you’ll miss proofs, here’s a proof related question that will boggle your mind: What is a reason that you can’t use in a proof to support a statement? Definition, postulate, yesterday’s theorems, or today’s theorems? The answer is upside down on the next slide.

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˙sɯǝɹoǝɥʇ s,ʎɐpɹǝʇsǝʎ

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