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Nice job Haley! Keep up mthe hard work. I appreciate your thorough explanations and reflections which indicate your growing understanding of the learning targets.

Mrs. Kudlov

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LT 12: I can recognize vertical angles. 

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LT 12: I can recognize vertical angles. 

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It is very easy to recognize vertical angles. We learned this when we were younger. The difference is now we can conclude that if they are vertical angles then they are congruent. If you have any more questions here is a video that helped me. It helped me remember that they will always be congruent and why.

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Geometry Proofs | Congruent Triangles & CPCTC

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Proof: Vertical angles are equal | Angles and intersecting lines | Geometry | Khan Academy

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This was really easy just remember the indicators of opposite rays.

1. 2 collinear rays

2. common endpoints

3. goes in different reactions

These makeup vertical rays.

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LT 10: I can apply the Subsitutution property.

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LT 10: I can apply the Subsitutution property.

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This was very easy because for the reason part pf the two column proof you only have to put substitution. This is such a relief from writing the whole theorem. To understand the substitution property you just have to understand when you use it in algebra. This worksheet really helped me which Mrs. Kudlov supplied.

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LT 9: I can apply the transitive properties of angles and segments.

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LT 9: I can apply the transitive properties of angles and segments.

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This was one of the easier theorems to understand. Basically you just have to understand that id A=B and B=C then A=C. Think of it as process of elimination. There two B's so the conclusion will be A and C. When given this problem use this theorem:

If two angles or segments are congruent to the same angle or segment then they are congruent to each other.

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LT 8: I can apply the multiplication and division properties of segments and angles. 

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LT 8: I can apply the multiplication and division properties of segments and angles. 

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These theorems were really easy to catch on to. The only problem was that they get really complicated when you add the subtraction properties and past theorems with them. You just have to take a breath and go through step by step. I had the problem with using the addition property when I needed to use the multiplication property. Just remember the indicators to use this theory.

1. Pair of congruent segments or angles

2. bisection or trisection

Also remember that if it goes from big to small then it is the division properties and if it goes small to big then it is the multiplication property.

Here's an example of the type of problem you'll see about multiplication an division properties.

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This unit was harder than the first unit. I did really well in the first until so for this one I just tried to keep up my work ethic. I continued to try on my homework and check hard. I learned that there was a lot more problems that I couldn't do this unit, but I always made sure that the next day I learned what to do in class. The hardest part was by far using all the different theorems because they get mixed up in your head. The best way to study is to go over the indicators for each theorem that way you know which one to use as soon as you see the given information.

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LT 7: I can apply the subtraction properties of segments and angles. 

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LT 7: I can apply the subtraction properties of segments and angles. 

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This learning target is the same as the addition property, but is with the subtraction theorem.

Subtraction theorem:

If congruent angles or segments are subtracted form congruent angles or segments then diff is congruent.

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Commutative & Associative Properties of Addition

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2.5 Addition and Subtraction Properties (Lesson)

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A lesson on the addition and subtraction properties and their application to geometry problems involving segments and angles

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LT 6: I can apply the addition properties of segments and angles. 

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LT 6: I can apply the addition properties of segments and angles. 

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This was easy once you got the hang of the theorems. The hardest part was applying them to algebraic problems. You just had to remember in which situations to subtract or add. It is all based on whether the sum is selected and a smaller angle or two smaller angles that would then equal the sum.

Here is the theorem which is reversible that we use for the addition problem:

If congruent segments or angles are added to congruent segments or angles then the sum is congruent.

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LT 5: I can prove angle congruent by using complementary and supplementary theorems. 

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LT 5: I can prove angle congruent by using complementary and supplementary theorems. 

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This is very simple once you get the hang of it. The biggest idea to remember for me was to include that the angles are adjacent during the theorems which means they are touching. As long as you remember the definitions in the previous learning target 3 and you use the following 2 theorems you will be able to prove congruent angles.

Comp Theorem: if two angles or adjacent segments sum is 90 degrees or a rt angle then it is a complementary pair.

Supp Theorem: if two angles or adjacent segments sum is 180 degrees or a st angle then it is a supplementary pair.

It also gets harder when there are multiple congruent angles that are also comp and supp. These are the following theorems that you would use:

If two angles comp or supp to the same angle or congruent angles then those angles are congruent.

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LT 3: I can recognize complemetary and supplementary angles. 

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LT 3: I can recognize complemetary and supplementary angles. 

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As long as you understand these definitions, you will be able use them for theorems and solving algebraic problems.

Complementary- the sums of the angles are 90 degrees or a right angle.

Supplementary- the sums of the angles are 180 degrees or a straight angle

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LT 2:  I can understand the concept of perpindicular. 

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LT 2:  I can understand the concept of perpindicular. 

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This was a very easy target because we have been hearing about perpendicular segments since middle school. All you have to understand is if two angles are perpendicular then they form right angles. This theorem is also reversible.