1. Know the postulates and theorems on pp. 770-2.
2. Go over the Chapter Reviews in the text.
3. Go through your quizzes.
4. There are extra practice problems starting on p.716. I would go over the following problems:
Chapter 1 (p.716): #18 - 37
#42 - 47
Chapter 2 : #1 - 28
Chapter 3: #1 - 13
#19 - 41
Chapter 4: #1 - 16
# 21 - 28
# 31-33
Chapter 5: #1 -9
# 13 - 20
# 33 - 42
Of course, you don't have to answer all of those but you can glance at each one and determine whether or not you know how to do it. If you're unsure, try the problem. Odd answers start on p.850.
I would say that most students had a little trouble with finding equations of lines. Make sure you know that material. Know how to find the slope of a line given two points, how to write the equation once you have a slope and how to tell whether lines are parallel or perpendicular.
If you have any questions, post them here or email me. I'll check in regularly.
Study!
Mrs. Dant
Saturday, December 18, 2010
Tuesday, December 7, 2010
Tips for New Proofs and Problems
You will have to use theorems from your notes for these problems. Problem #11 tells you to use Theorem 24, which is not in your book. Here it is:
Thm 24 - If two points are each equidistant from the endpoints of a segment, then the two points determine the perpendicular bisector of that segment.
Another useful theorem:
Thm 25 - If a point is on the perpendicular bisector of a segment, then it is equidistant from the endpoints of that segment.
TIPS:
4.) In order for PQ to be the perp. bis. of RS, P and Q must be equidistant from R and S. That means, PR = PS and QR = QS. You don't have to go from congruent segments to equal distances. If you have that PR congruent to PS, that means equidistant. Tip - draw radii. (4 step proof)
5.) You do not have to state that E is on the perpendicular bisector. Let's all agree that it is! Use Thm 25. (5 step proof)
6.) Use subtraction property. (3 step proof, with semicolons)
7.) Just answer it.
8.) Same.
9.) Again.
10.) a.) Ignore angle EBC. You can set up an equation with the two expressions. Make sure you use the right equation. Don't assume that segment AB is perpendicular to CD when you write the eq'n.
b.) Use the solution for x from part a to answer this.
11.) Given: PQ = PR (actually congruent but I don't have that key)
SQ bisects <PQR
______________ (you fill in this one)
Prove: PS perp QR
I used the Division property here, followed by another common theorem. You'll also use Thm. 24. (about 6 steps)
12.) There's a hint already there. To tell which segment is the perp. bisector, consider one point at a time. Once you find two points that are each equidistant to the endpoints of a segment, "connect them" to form the perp. bisector.
13.) Don't neglect to use the transitive property so that you can get all four segments congruent.
14.) Once you have the segments perpendicular, choose one angle to be right and then you can conclude the Prove.
15.) Number four acute angles. (Not the right angles.) You'll have to use Subtraction and 'If angles, then sides.' Theorem 24 is handy too.
16.) Make sure you DO NOT use congruent triangles. To prove something is an altitude, the step before that must include perpendicular segments. (4 step proof)
17.) Ok, on this one, you can prove one set of triangles to be congruent. Then use Thm 24.
18.) Again, what's the definition of altitude? Remember that the area of a triangle is 1/2 (base)(height). But, in a right triangle, the base and the height are the legs.
19.) a.) Figure out the coordinates of C and T.
b.) When you rotate point E 90 degrees clockwise, its coordinates are easier to figure out. (Only 1, 12, -1, or -12) are possible. Use that for a simpler sample and follow the pattern.
Thm 24 - If two points are each equidistant from the endpoints of a segment, then the two points determine the perpendicular bisector of that segment.
Another useful theorem:
Thm 25 - If a point is on the perpendicular bisector of a segment, then it is equidistant from the endpoints of that segment.
TIPS:
4.) In order for PQ to be the perp. bis. of RS, P and Q must be equidistant from R and S. That means, PR = PS and QR = QS. You don't have to go from congruent segments to equal distances. If you have that PR congruent to PS, that means equidistant. Tip - draw radii. (4 step proof)
5.) You do not have to state that E is on the perpendicular bisector. Let's all agree that it is! Use Thm 25. (5 step proof)
6.) Use subtraction property. (3 step proof, with semicolons)
7.) Just answer it.
8.) Same.
9.) Again.
10.) a.) Ignore angle EBC. You can set up an equation with the two expressions. Make sure you use the right equation. Don't assume that segment AB is perpendicular to CD when you write the eq'n.
b.) Use the solution for x from part a to answer this.
11.) Given: PQ = PR (actually congruent but I don't have that key)
SQ bisects <PQR
______________ (you fill in this one)
Prove: PS perp QR
I used the Division property here, followed by another common theorem. You'll also use Thm. 24. (about 6 steps)
12.) There's a hint already there. To tell which segment is the perp. bisector, consider one point at a time. Once you find two points that are each equidistant to the endpoints of a segment, "connect them" to form the perp. bisector.
13.) Don't neglect to use the transitive property so that you can get all four segments congruent.
14.) Once you have the segments perpendicular, choose one angle to be right and then you can conclude the Prove.
15.) Number four acute angles. (Not the right angles.) You'll have to use Subtraction and 'If angles, then sides.' Theorem 24 is handy too.
16.) Make sure you DO NOT use congruent triangles. To prove something is an altitude, the step before that must include perpendicular segments. (4 step proof)
17.) Ok, on this one, you can prove one set of triangles to be congruent. Then use Thm 24.
18.) Again, what's the definition of altitude? Remember that the area of a triangle is 1/2 (base)(height). But, in a right triangle, the base and the height are the legs.
19.) a.) Figure out the coordinates of C and T.
b.) When you rotate point E 90 degrees clockwise, its coordinates are easier to figure out. (Only 1, 12, -1, or -12) are possible. Use that for a simpler sample and follow the pattern.
Thursday, November 25, 2010
Five Challenging Proofs
OK guys, these are tough, particularly #17 and #18. Once you get them, however, they might become your favorite proofs of the year! I don't want to give away too much because I'll be robbing you of the joy that comes from figuring them out. (Yes, I did say 'joy'.)
OK, here goes... For each of these, you'll have to prove at least two sets of triangles congruent. Use the Givens and the diagrams to get the first sets. Then, carefully consider each part of those triangles until you find the parts that can be used ~ with CPCTC ~ to get the next set of triangles. Always keep your eye on which parts you're aiming for.
USE COLORED PENCILS TO OUTLINE TRIANGLES ONCE YOU HAVE THEM CONGRUENT.
14.) Subtraction property is quicker than the Congruent Complements Theorem. Start with the small triangles.
15.) Draw. I used SSS and ASA.
16.) This proof took me 11 steps and I did not combine Givens. From the Division Property, you'll need to get two conclusions. You can list them both right away and then use Restates if/when you need to. 'If angles, then sides' and its converse will be very useful.
17.) OK, now comes the real fun!! The first time through, this took me 19 steps but I was convinced I could be more efficient. Sure enough, I had neglected to use the obvious vertical angles. Fixing that saved me four steps! I used THREE sets of triangles on this one so that means I have six triangles listed in my proof. Two of them are RPU and SUQ (which are not congruent to each other, of course). I won't spoil the ending.
Some of my reasons include Subtraction property, Reflexive property, CPCTC, SAS and Vertical angles are congruent. (**Note: For the reflexive property, you can call the same angle by two different names. I did this for the first set of triangles.)
You might want to use backward braces for those statements which give you more than one important conclusion. For example, the congruence statement for triangle RPU gave me two parts that I wanted to use so I used a backward brace there.
18.) As soon as I used a colored pencil, I figured this one out. It's tough because there are so many triangles that you could try. This one took me 14 steps and two sets of triangles. The first set of triangles is easy because the Givens are all you need. Use a colored pencil to outline those.
Then, you move on to the next set of triangles. I chose a set that included angle DAC. You'll be able to use SAS for those. (Number the small angles by vertex D.)
Do not use any triangles that do not have a labeled point at each vertex.
Lastly, 'if sides, then angles' will come in handy with segments AD and BD. A little subtraction is all you'll need to close it up.
TO RECAP THE GRADING:
If you seriously attempt all five, you will be eligible for an A. If you try 3 or 4, you can earn a B. If you try 1 or 2, you can earn a C. I will be grading for content from there.
Have fun with these... every bit as enjoyable as a riddle, brain teaser or Sudoku puzzle. = + )
Happy Thanksgiving and go Cadets!!!!
OK, here goes... For each of these, you'll have to prove at least two sets of triangles congruent. Use the Givens and the diagrams to get the first sets. Then, carefully consider each part of those triangles until you find the parts that can be used ~ with CPCTC ~ to get the next set of triangles. Always keep your eye on which parts you're aiming for.
USE COLORED PENCILS TO OUTLINE TRIANGLES ONCE YOU HAVE THEM CONGRUENT.
14.) Subtraction property is quicker than the Congruent Complements Theorem. Start with the small triangles.
15.) Draw. I used SSS and ASA.
16.) This proof took me 11 steps and I did not combine Givens. From the Division Property, you'll need to get two conclusions. You can list them both right away and then use Restates if/when you need to. 'If angles, then sides' and its converse will be very useful.
17.) OK, now comes the real fun!! The first time through, this took me 19 steps but I was convinced I could be more efficient. Sure enough, I had neglected to use the obvious vertical angles. Fixing that saved me four steps! I used THREE sets of triangles on this one so that means I have six triangles listed in my proof. Two of them are RPU and SUQ (which are not congruent to each other, of course). I won't spoil the ending.
Some of my reasons include Subtraction property, Reflexive property, CPCTC, SAS and Vertical angles are congruent. (**Note: For the reflexive property, you can call the same angle by two different names. I did this for the first set of triangles.)
You might want to use backward braces for those statements which give you more than one important conclusion. For example, the congruence statement for triangle RPU gave me two parts that I wanted to use so I used a backward brace there.
18.) As soon as I used a colored pencil, I figured this one out. It's tough because there are so many triangles that you could try. This one took me 14 steps and two sets of triangles. The first set of triangles is easy because the Givens are all you need. Use a colored pencil to outline those.
Then, you move on to the next set of triangles. I chose a set that included angle DAC. You'll be able to use SAS for those. (Number the small angles by vertex D.)
Do not use any triangles that do not have a labeled point at each vertex.
Lastly, 'if sides, then angles' will come in handy with segments AD and BD. A little subtraction is all you'll need to close it up.
TO RECAP THE GRADING:
If you seriously attempt all five, you will be eligible for an A. If you try 3 or 4, you can earn a B. If you try 1 or 2, you can earn a C. I will be grading for content from there.
Have fun with these... every bit as enjoyable as a riddle, brain teaser or Sudoku puzzle. = + )
Happy Thanksgiving and go Cadets!!!!
Monday, November 1, 2010
A Few New Tips - CPCTC
Hi everybody,
I hope that triangle proofs are becoming easy for you. I'm confident that you all can handle this next proof packet with little trouble. I'll include tips for those that are more than a few steps. You're on your own for the short ones - that is, unless you ask me about them at school.
4. Only 2/3 of the trisection in the Given is important. Number the right angles 1 and 2. Don't forget to go through the pedantic steps with perpendicularity. (Right, Kory?)
5. Start with the midpt. step.
6. Easy. If you choose to put numbers into angles, change the Givens to reflect the new angle names.
7. Remember your formulas?
8. This is not a system of equations, just three separate eqn's. To isolate a variable that is under a square root symbol, you have to do the opposite operation to both sides. Check your answers!
9. Easy. We've seen this diagram in the last section. Five-step proof.
10. Start with MP = RO. That alone leads to a needed step.
11. Shaded region problems? Imagine the whole diagram is shaded and find area. Then "cut out" the appropriate section.
12. Hmmm... midpt. is used twice. Even though this isn't a proof, I'd like to see the diagram drawn and marked. I'll leave it up to you to decide if this is a system of equations or just two equations to be solved separately. Remember this general rule: to solve for n variables, you need n equations. (There are exceptions.)
13. Remember our special definition of right angles? It was that if two angles form a right angle, then they are complementary.
14. Very similar to #13. This might be the last time you use the Linear Pair Postulate.
15. Turn around the second Given if you want to see the Subtraction Property work smoothly. Don't use extra steps.
16. I'll allow a double-transitive in one step. Make it look like a triple play.
17. With overlapping triangles, you should choose which ones to go for. You can complete this proof with the two smallest triangles or the overlapping ones. Which way is more efficient? The Prove statement usually leads you. Six-step proof.
18. Looks much harder than it is. Should I even mention the obvious triple play? I would suggest starting with those statements and reaching the first conclusion. Under the triple play, put the other Given that needs to be worked on a little. That way, your 3 important triangle statements should line up nicely.
19. Why is this here? Trust your common sense.
20. The first proof that goes beyond CPCTC - not very far beyond, though! Yes, you can draw them in.
21. OK, this is the last time you use LPP. With those included into one, I get nine steps.
18.
I hope that triangle proofs are becoming easy for you. I'm confident that you all can handle this next proof packet with little trouble. I'll include tips for those that are more than a few steps. You're on your own for the short ones - that is, unless you ask me about them at school.
4. Only 2/3 of the trisection in the Given is important. Number the right angles 1 and 2. Don't forget to go through the pedantic steps with perpendicularity. (Right, Kory?)
5. Start with the midpt. step.
6. Easy. If you choose to put numbers into angles, change the Givens to reflect the new angle names.
7. Remember your formulas?
8. This is not a system of equations, just three separate eqn's. To isolate a variable that is under a square root symbol, you have to do the opposite operation to both sides. Check your answers!
9. Easy. We've seen this diagram in the last section. Five-step proof.
10. Start with MP = RO. That alone leads to a needed step.
11. Shaded region problems? Imagine the whole diagram is shaded and find area. Then "cut out" the appropriate section.
12. Hmmm... midpt. is used twice. Even though this isn't a proof, I'd like to see the diagram drawn and marked. I'll leave it up to you to decide if this is a system of equations or just two equations to be solved separately. Remember this general rule: to solve for n variables, you need n equations. (There are exceptions.)
13. Remember our special definition of right angles? It was that if two angles form a right angle, then they are complementary.
14. Very similar to #13. This might be the last time you use the Linear Pair Postulate.
15. Turn around the second Given if you want to see the Subtraction Property work smoothly. Don't use extra steps.
16. I'll allow a double-transitive in one step. Make it look like a triple play.
17. With overlapping triangles, you should choose which ones to go for. You can complete this proof with the two smallest triangles or the overlapping ones. Which way is more efficient? The Prove statement usually leads you. Six-step proof.
18. Looks much harder than it is. Should I even mention the obvious triple play? I would suggest starting with those statements and reaching the first conclusion. Under the triple play, put the other Given that needs to be worked on a little. That way, your 3 important triangle statements should line up nicely.
19. Why is this here? Trust your common sense.
20. The first proof that goes beyond CPCTC - not very far beyond, though! Yes, you can draw them in.
21. OK, this is the last time you use LPP. With those included into one, I get nine steps.
18.
Wednesday, October 13, 2010
Last Set of Tips - Multiplication and Division Properties
Using the Multiplication and Division Properties in Proofs
1. Look for a double use of the word midpoint or trisect or bisect in the Givens.
2. Use the Multiplication property when the segments or angles in the Prove are larger than those in the Given.
3. Use the Division property when the segments or angles in the Prove are smaller than those in the Given.
(taken from Geometry for Enjoyment and Challenge, McDougal & Littell, Evanston, 1991, p.90)
Read and reread the above steps, until you understand them!!
Tips:
#1, 3, 4, 5) Two-step proofs.
#6) Two-stepper but be careful.
#8)Why is this in there?! Two steps, ridiculously easy!!
#11) Fun proof! Ignore the first given for a moment. Use the other three with the Mult. or Division property. You should be able to conclude something about segment ZX. Now use the first Given and you should have a good proof. Altogether, four steps if you combine the three related Givens into one step.
#12) Even more fun!! I don't want to spoil it. You should see a supplementary double-play; that's where I would start. By combining similar steps, I would get four steps.
#13) Start with linear pairs, move into a supplementary triple-play. With that conclusion, use either the Mult. or Div. property. Again, combining similar steps with semi-colons, I would get six steps.
#16) The reasons I would use for this proof, out of order, would include Division Property, LPP, Addition Property, Given, Diagram, Congruent Supplement Theorem, and Restate (you'll use <1 = <2 twice so you can call it Given or Restate). Hardest proof in the packet. Eight steps.
1. Look for a double use of the word midpoint or trisect or bisect in the Givens.
2. Use the Multiplication property when the segments or angles in the Prove are larger than those in the Given.
3. Use the Division property when the segments or angles in the Prove are smaller than those in the Given.
(taken from Geometry for Enjoyment and Challenge, McDougal & Littell, Evanston, 1991, p.90)
Read and reread the above steps, until you understand them!!
Tips:
#1, 3, 4, 5) Two-step proofs.
#6) Two-stepper but be careful.
#8)Why is this in there?! Two steps, ridiculously easy!!
#11) Fun proof! Ignore the first given for a moment. Use the other three with the Mult. or Division property. You should be able to conclude something about segment ZX. Now use the first Given and you should have a good proof. Altogether, four steps if you combine the three related Givens into one step.
#12) Even more fun!! I don't want to spoil it. You should see a supplementary double-play; that's where I would start. By combining similar steps, I would get four steps.
#13) Start with linear pairs, move into a supplementary triple-play. With that conclusion, use either the Mult. or Div. property. Again, combining similar steps with semi-colons, I would get six steps.
#16) The reasons I would use for this proof, out of order, would include Division Property, LPP, Addition Property, Given, Diagram, Congruent Supplement Theorem, and Restate (you'll use <1 = <2 twice so you can call it Given or Restate). Hardest proof in the packet. Eight steps.
Hints for Addition & Subtraction Properties
Here are a few more tips for the proof packet:
# 3, 4, 5) Two-step proofs (the only thinking required is Reason #2 and it'll either be the Addition Property or Subtraction Property)
# 7) Same as above but you have to come up with your own Conclusion (or Prove) statement.
# 8) Old stuff - If you combine all Givens, it'll be only 2 steps!
#10) The way I did this in class is actually the old way... there's an easier method. I get 5 steps, the last one using the subtraction property.
Remember, to use the subtraction property, you only need this format:
1. Get two "big" angles congruent.
2. Get two "little" angles congruent - as long as they're inside the big ones.
3. Put a brace around those two steps and conclude that the remaining little angles are congruent.
#11) Four steps and the reasons are 1. Given 2. Definition of bisector 3. Given 4. ______________
#12) Start with the trisection. What does that mean?? It means that the segment is divided into three congruent segments. Two of them are important in this problem. You need to add them to other segments. Should take four steps.
#13) Almost identical to #10. Remember that all right angles are congruent. If you start with the perpend. step, you should follow that with a right angle step. Throw in the other right angle statement that you're given and then state that they are congruent. It will take two more steps to reach the Prove. In total, six steps.
# 3, 4, 5) Two-step proofs (the only thinking required is Reason #2 and it'll either be the Addition Property or Subtraction Property)
# 7) Same as above but you have to come up with your own Conclusion (or Prove) statement.
# 8) Old stuff - If you combine all Givens, it'll be only 2 steps!
#10) The way I did this in class is actually the old way... there's an easier method. I get 5 steps, the last one using the subtraction property.
Remember, to use the subtraction property, you only need this format:
1. Get two "big" angles congruent.
2. Get two "little" angles congruent - as long as they're inside the big ones.
3. Put a brace around those two steps and conclude that the remaining little angles are congruent.
#11) Four steps and the reasons are 1. Given 2. Definition of bisector 3. Given 4. ______________
#12) Start with the trisection. What does that mean?? It means that the segment is divided into three congruent segments. Two of them are important in this problem. You need to add them to other segments. Should take four steps.
#13) Almost identical to #10. Remember that all right angles are congruent. If you start with the perpend. step, you should follow that with a right angle step. Throw in the other right angle statement that you're given and then state that they are congruent. It will take two more steps to reach the Prove. In total, six steps.
Sunday, October 10, 2010
Video for Constructions
Hi Everybody,
I found a video on You Tube that should be helpful for you. It covers most of the basic constructions that you've learned in class. The man in the video does segment construction differently than we do -- he doesn't use a working line. The particular construction that you should focus on is parallel line using alternate interior angles.
About the trapezoid construction, you can simply construct the parallel line exactly the way he does. I'll be giving you the lenghts of the bases by showing two segments. After you construct the two parallel lines, you can make each of them congruent to "my" segments and then simply connect the endpoints. You should have no trouble with that.
Enjoy the video. Sorry I couldn't introduce you to my dog but I don't have access to a video camera at the moment. Hope you're enjoying your weekend.
Mrs. Dant
Click here: Geometric Construction
I found a video on You Tube that should be helpful for you. It covers most of the basic constructions that you've learned in class. The man in the video does segment construction differently than we do -- he doesn't use a working line. The particular construction that you should focus on is parallel line using alternate interior angles.
About the trapezoid construction, you can simply construct the parallel line exactly the way he does. I'll be giving you the lenghts of the bases by showing two segments. After you construct the two parallel lines, you can make each of them congruent to "my" segments and then simply connect the endpoints. You should have no trouble with that.
Enjoy the video. Sorry I couldn't introduce you to my dog but I don't have access to a video camera at the moment. Hope you're enjoying your weekend.
Mrs. Dant
Click here: Geometric Construction
Thursday, October 7, 2010
Sample Proofs for Packet
With the new theorems discussed in class, your proof packets should become much easier. Below is a link that shows how you should use the new addition and subtraction properties. Remember, they're not very different from the addition and subtraction properties we used in algebra proofs.
Here's the reasoning of these properties:
1. Start with two equal "things"
2. Add/Subtract the same thing (or equal things) to both
3. Get equal results
Samples of New Addition/Subtraction Properties - Basic
Here's the reasoning of these properties:
1. Start with two equal "things"
2. Add/Subtract the same thing (or equal things) to both
3. Get equal results
Samples of New Addition/Subtraction Properties - Basic
Sample Proofs on Parallel Lines
Hi everybody,
If you click on the Sample link below, you should be taken to a Google Docs site to which I've uploaded two sample proofs. I know that you have samples in your notes but I wanted to include these below for an extra resource.
Sample Flow Proofs - Chapter 3
I'll add to this periodically. Don't hesitate to email me with questions.
Mrs. Dant
If you click on the Sample link below, you should be taken to a Google Docs site to which I've uploaded two sample proofs. I know that you have samples in your notes but I wanted to include these below for an extra resource.
Sample Flow Proofs - Chapter 3
I'll add to this periodically. Don't hesitate to email me with questions.
Mrs. Dant
Tuesday, October 5, 2010
Proof Tips I
Hi Boys,
For the first page of proofs, you should be focusing on the supplementary/complementary "double and triple plays." Remember, if you can get two supps or two comps into a proof, you'll probably end up with congruent angles.
Example: 1. < A comp < B
2. < C comp < D -> 4. < B = < C
3. < A = < D
The reason for statement 4 would be: Complements of congruent angles are congruent.
Page 1
- all double or triple plays
Page 2
- #9 is a simple warm-up to remind you what bisectors are --- 2 step proof!
- #10 should have a reason which states If two lines are perpendicular, they intersect to form right angles.
- #11 - Start the proof with GJ bisects <FGH. It will line up nicer and it's OK to have a triple play written "upside down."
- #16 Remember that definitions are reversible. The definition of bisects is this: If a ray bisects an angle, then it divides the angle into two congruent angles. Use that definition? Reverse it? Both????
Don't give up easily - THINK HARD!!!
For the first page of proofs, you should be focusing on the supplementary/complementary "double and triple plays." Remember, if you can get two supps or two comps into a proof, you'll probably end up with congruent angles.
Example: 1. < A comp < B
2. < C comp < D -> 4. < B = < C
3. < A = < D
The reason for statement 4 would be: Complements of congruent angles are congruent.
Page 1
- all double or triple plays
Page 2
- #9 is a simple warm-up to remind you what bisectors are --- 2 step proof!
- #10 should have a reason which states If two lines are perpendicular, they intersect to form right angles.
- #11 - Start the proof with GJ bisects <FGH. It will line up nicer and it's OK to have a triple play written "upside down."
- #16 Remember that definitions are reversible. The definition of bisects is this: If a ray bisects an angle, then it divides the angle into two congruent angles. Use that definition? Reverse it? Both????
Don't give up easily - THINK HARD!!!
Wednesday, September 15, 2010
Tuesday, September 14, 2010
Welcome to our class blog!
Hi geometers,
This blog is intended to help you navigate through geometry class when the textbook isn't quite enough. I will post tips and links to other sites that should clear things up for you. If you have a Google account and you sign on to this page, you can post messages too. Feel free to post a message for me or for your classmates, as long as it pertains to geometry.
I'll be posting some construction tips before Friday's quiz. Check back soon.
Mrs. Dant
This blog is intended to help you navigate through geometry class when the textbook isn't quite enough. I will post tips and links to other sites that should clear things up for you. If you have a Google account and you sign on to this page, you can post messages too. Feel free to post a message for me or for your classmates, as long as it pertains to geometry.
I'll be posting some construction tips before Friday's quiz. Check back soon.
Mrs. Dant
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