Project #2 Peer Feedback

	A	B	C	D	E	F	G
1	Presenter's name	Topic or theorem name	Content knowledge	Pedagogy	Technology	Write the statement of the theorem in your own words	Feedback

2	Armando	Desargues theorem	2	2	3	if we have two triangles ABC and A'B'C' in a plane such that they are different from each other and their corresponding sides are not parallel, if we extends their three sides, every two corresponding sides of these two triangles will intersect at a point and that called the Axial perspectivety. and the concurrence of these three lines called the central perspectivity. and the converse is true.	Good job but It would have been better if Armando gave more information on the theorem on his open lab post rather then just writing what the theorem is.
3	Armando	Desargues' Theorem	3	4	4	If two triangles are perspective from a point, then they are perspective from a line; if two triangles are perspective from a line, then they are perspective from a point.	I like how Amando defined the perspector and the perspectrix of two triangles. Also, he made questions for us to think about the theorem. I used the drag test to change the shape of the triangles and I could not find the perspector and the perspectrix kind of not existing any more. So, does it mean that for any two triangle we can at most find one perspector and one perspectrix. What are the conditions for these two triangles. I found this topic very interesting because it causes many questions to me when I really think about it. For interpreting this topic, Amando did a very good job. For further exploration, we need some formal proof when we need to really understand it deeply.
4	Armando	Desargues's Theorem	3	3	4	Two triangles are perspective from a point if there are perspective from a line. But the points have to be collinear on the perspectrix. If not, then the two triangles are not perspective from a point.	I thought it was very clever to break down the theorem.
5	Armando	Desargue's Theorem	4	4	4	The case where you have two triangles where three lines forming from the corresponding vertices of each triangle meets at a point called the Perspector. Then the same triangles with lines forming from the sides of each triangle and make new points at which they meet with the line formed from the corresponding sides of the other triangle, the three new points are formed and the line connecting each new point shows linearity and this line is called the Perspectrix. This means that they are perspective from a point in which they are perspective from a linear line, hence the Desargue's Theorem.	At first this seemed like a complicated theorem to explain, but I understand it through your demonstration and definitions.
6	Armando	Desargues's Theorem	4	4	4	Two triangles ABC and A'B'C' are congruent by point D.	I enjoyed how each of the slides showed a step by step guide on how to recreate the theorem and also using the different colors to show the lines to help make it easier to see what the words are saying. The weakness of the presentation is that you could have defined words for the words that students are having trouble with while reading the sentences that you wrote.
7	Armando	Desargues Theorem	4	4	4	In the special case of two triangles whose points are collinear prove that the perspective of three intersecting points of the corresponding sides form the perspective line .	This was a great worksheet. good use of check boxes for questions. Providing the necessary information or in your words ( def'n) really helped clarify each characteristic of the triangles perspective.
8	Armando	Desargues' Theorem	3	3	4	I understand that the Desargues' Theorem is when the 2 triangles are perspective from a point that is perspective from a line. however to me one of the triangle seems to be the reflexion of the other one.	Overall the project is okay. but when using the drag test, point D need a little adjustment. And also he needs to be careful with the difference between "Their and there."
9	Eve	Law Of Sines	4	4	4	If a Triangle has a circumradius R, then two times the radius is equal to a side of the triangle divided by sine of the angle opposite of the side.	The color coating really made it helpful for me. Amazing job. I appreciate you.
10	Eve	Extended Law of Sines	4	4	3	The circle formed around the vertices of a triangle will have a radius that is approximately double the ratios of sides of the triangle and the angles of each connecting line.	I was able to follow the presentation, but it got a little cluttered fast. But, I do understand that there was a lot of explaining you wanted to do in such short time. I like how you have the option to clear space.
11	Eve	Extended Law of Sines	4	4	4	Is a way to help find the lengths and the measure of the angles of a triangle ABC.	The presentation was detailed step by step on how the theorem works and the presentation was also through. The presentation also showed the measures of the different sides and angles to help me better understand what was going on in the theorem.
12	Eve	The Extended Law of Sines	4	4	4	In a triangle with circumscribed circle around, there is a relationship between the radius of the circle r and the ratio of the side over the sine of the opposite angle. So, BC/sinA) = AC/sin(B) = AB/sin(C)=2r	This is my favorite worksheet. :) I love how Evelin used the color background for her textboxes. The idea of the theorem is very and understandable. Evelin used the drag test very well to show the theorem. Excellent.
13	Eve	Extented law of sines	3	3	2	If we draw a triangle inside a circle( circumscribed circle) with the center o( circumscribed to circle) such that each three vertices are in the circle then the sum of each the measure of segment in the triangle over the sin of the opposite angle (segment/opposite angle) equals the radius R/2	Eveline did an extinguished job on the dynamic worksheet, her instructions from drawing a center O to the end of the worksheet were explicative and on the point of each step.
14	Eve	Extended law of Sines	3	3	3	Use the law of sine to show that for any triangle ABC the angles of the given vertices can be .	As always your ability to explain things are uncanny. I now see how the law of sine can also be used for determining the radius of vertices with respect to the circumcenter.
15	Eve	Extended Law of Sines	3	4	4	Extended Law of Sines is when the proportionality of the segments to the sine of the angle is equal to each other and also equal to 2R	The content is okay and everything is defined.
16	Gary	Morley Theorem	4	4	4	the closer intersection of rays to its original vertex of an angle that has been trisected form an equilateral triangle.	I would have liked to hear what a trisection was, I did not follow where the A' and A'' came from. other than that, Great job, it was very clear and I could understand what you did very well.
17	Gary	Morle's Theorem	4	3	4	In a triangle ABC, creating the angle trisectors of angles of angles A, B, and C. The intersections of the adjacent angle trisectors will form an equilateral triangle.	The implementation of Geogebra was well performed, but the explanation of adjacent angle trisectors was a little confusing. There are many intersections for sure, and Gary's use of "the closest to the side" was a little misleading because it means that the distance is shortest. However, there is a case that the other intersection was actually closer to the side than the intersection of adjacent angle trisector. In order to make this clear, I suggest Gary to be consistent with his words. For example, keep all the sentences like: the adjacent angle trisectors of angle A and angle B intersect at D, the adjacent angle trisectors of angle A and angle C intersect at E, and the adjacent angle trisectors of angle B and angle C intersect at F. Points D, E, and F form an equilateral triangle.
18	Gary	Morley Theorem	3	3	2	The three points of intersection of the trisectors form an equilateral triangle.	What I liked about your presentation was that you gave a definition for each term but I feel like you could've explained a bit more of why that is happening. You could've also used other geogebra features to help you explain the theorem more.
19	Gary	Morley's Theorem	4	4	3	The interior angles of a triangle is split up into three equal parts called a trisector and the rays extending from each part intersects at points forming a new triangle which will always be an equilateral triangle.	I love how you ask the question about what type of triangle forms from the intersecting points, but I wish you would've gave a little bit more detail about the theorem or your diagram, like what happens to the equilateral triangle that is formed when the degree of the original triangle's angle changes.
20	Gary	Morley theorme	3	3	3	Morley's theorem state that for any triangle the three point of the intersection of the adjacent angle trisectors( an Array that divides the angle into three equal part) form an equilateral triangle	I think that Gary did a good job, the worksheet is clear and organized.
21	Gary	Morely Theorem	3	3	4	Prove that for any arbitrary triangle when you divide it into three parts. The trisecting parts form an equilateral triangle.	I think incorporating a little more explanation would be helpful. Do we really need the angle measurements for triangle ABC?
22	Gary	Morley Theorem	3	3	4	Based the quick definition on openlab and project the morley Theorem the triangle that is formed by the ray from the trisection angle.	A little bit more definition of the theorem and a quick explain of the angle is missing.
23	Jodel	Theorem of Menelaus	3	2	3	Honestly, I can't come up with a conclusion of a theorem with this worksheet.	I found so many flaws with this, a few to pick out, The theorem states F is on Segment AC, but on the worksheet, it is not. Second, I have no clue what L, M, and K represent. Third, the ratio represented in the worksheet only shows the ratio for one triangle, not one subject to the drag test. The technology was used well, but could have added a few more items to make it fit a bit better.
24	Jodel	The Theorem of Menelaus	4	4	4	It doesn't matter how you drag the segments, the final result will always be 1.	Overall, I found your presentation to be easy to follow and easy to understand. I would just recommend that you use the values of the segment into the theorem. As you drag the triangle around, It would be helpful to see that although the values change, it is still equal to 1.
25	Jodel	Theorem of Menelaus	4	3	2	The line that crosses points on the sides of a triangle is called Menelaus. The proportionality of each line created is equal to 1.	I'm a little confused about your proportions, because I thought that you were comparing the ratio of each line segment for an instance line AD/DB BE/EC, so why FC/AF and not AC/CF. I'm a little confused.
26	Jodel	Menelaus	4	4	4	In triangle ABC, a transversal line crosses BC, AC and AB at the points of D, E and F respectively, with D, E, and F.	The presentation was well written with detailed explanation on what each part of the theorem was saying. The weakness of the presentation was at the beginning when you just place the points on the triangle and the line connecting the points and didn't explain why were you connecting them from the start.
27	Jodel	Menelaus Theorem	3	2	2	For a triangle ABC, let points D, E, and F on the sides of AB, BC, and AC accordingly. We call the three points Menelaus points. They are collinear if and only if (AD/DB)(BE/EC)(CF/FA)=1.	For the point F, Jodel should show us that it is the intersection of lines DE and AC. So, it is clear to us that points D, E, and F are collinear since two points make a line. For the wording, I suggest Jodel used the desktop version to adjust the textbox because when we play it, the textboxes will not be floating around. It was not easy to understand the theorem when the statements are not organized. Also, for the formula, if it can be modified so we can verify the statement when we use the drag test, then it will be a lot better.
28	Jodel	Menelaus Theorem	3	3	3	For any given line that crosses three sides of a triangle one of them need to be extended . the product of three segments is equal to the the product of three others.	Clear and concise. Use of geogebra worked really well here.
29	Jodel	The Menelaus Theorem	2	3	3	Menelaus theorem say that if we draw a transverse that crosses any tow line segment in a triangle ABC then this proportionality AD/DB=DE/EC=FC/EF is equal to -1 such that D,E and F lie respectively in AB, BC, and AC.	I think Jodel did a great job but I think proportionality of the line segments is equal to -1 not 1.
30	Josiel	Napoleans Theorem	4	4	4	When you have one equilateral triangle and have others same triangles on the side. Then finding the medians of the triangles I can see that the medians can create another equilateral triangle.	The student worksheet was well written and the worksheet picture was detailed with words on what was happening. The weakness of the worksheet was that the lines were blocking some of the words.
31	Josiel	Napoleans Theorem	4	3	4	For a given triangle, we can create three different equilateral triangles on the sides of the triangle. Then, we create the centroids of the equilateral triangles. We connect the three centroids and we will form another equilateral triangle. In order to find the Napolean point, we need to connect the vertices from the original given triangle to the opposite vertices of the created equilateral triangle. The intersection is the Napolean point.	Josiel has a very good understanding of Geogebra command. He used the check box command, which is great to see. Just one clarification, I think for the created equilateral triangle GHI, the intersection being created is not the centroid. So, we should remove the Medians checkbox for the triangle GHI. The last suggestion is changing the color for the lines to make some important segments or medians clear. Overall, the work is fantastic.
32	Josiel	Napolean Theorem	4	4	4	If you connect 3 equilateral triangles where one of its sides is a side of another triangle, and you find all 3 centroids of those equilateral triangles, then the centroids will itself create an equilateral triangle.	Those check boxes were a life saver!! sheesh!!! Colors really helped me understand it. Really great job!!
33	Josiel	Napoleans theorem	3	4	4	if we start with any triangle ABC and we draw the three equilateral triangles taking the measure of the original triangle sides followed by drawing all the centroids of the three equilateral triangles, these three centroids form a triangle that is called the napoleons triangle and its always an equilateral triangle no matter what original triangle ABC we started with, in addition centroid of the resulting triangle( napoleans triangle) is called the napoleons point	I think this is the most complicated theorem of them all and Josiel did a very good job on illustrating it, the steps on the geogebra worksheet are clear. every step matches its drawing perfectly. the fact that he included the check boxes made it easy to say through triangles, because at on point there were a lot of lines, and those check boxes mask unwanted lines so we can focus on the resulting triangle.
34	Josiel	Napoleons Theorem and the Napoleon Point	4	4	4	The Napoleon Point is formed when the medians of the triangle (the triangle that was formed from the three centroids of the three equilateral triangles) intersect. That same triangle will always be equilateral regardless of the size from the original triangle.	Some of the strengths from this worksheet was definitely the check boxes. It helped me see the image clearer and it was more organized. The only suggestion I can recommend is providing the text before performing the action. Overall, I found this worksheet really easy to understand.
35	Josiel	Napoleon's Theorem	4	4	4	Using equilateral triangle ABC, create three more attached by each leg on the triangle creating a centroid in the middle of each new triangle will create points to form another equilateral triangle GHI. This triangle will always remain an equilateral triangle when triangle ABC changes its position. The centroid of triangle GHI is called Napoleon's point.	Your demonstration of the theorem was simple and understandable.
36	Josiel	Napoleans Theorem	3	3	4	Given any triangle ABC prove that when created three equivalent triangles on the vertices ABC the centroids of each triangle create three points called the napolean point that constructs a new triangle that will maintain its equilateral sides.	I thought this was an interesting theory the use of the check boxes really helped clarify what i was observing. I did notice your text boxes starting staggering to the left did you mean to do this on purpose to solidify the claim ?
37	Josiel	Napoleans Theorem and the Napoleans points	3	3	3	Based on the project, the Napolean theorem is the triangle that forms the centroid of the previous triangles.	It could execellent if he did not forget to explain what is Napoleans points.
38	Josiel	Napoleans Theorem and the Napolean Point	3	4	4	Napoleon 's theorem stats that if we construct equilateral triangles on the sides of any triangle, the centers of those equilateral triangle will from an equilateral triangle, we call it Npoleon triangle.	I think Josiel's dynamic worksheets shows steps of how to find a the Napolean Points very claerly.But the only thing that need to improve is the notes should be more organized
39	Luis	The medial triangle	3	3	4	A medial triangle (in this case triangle ABC is the medial triangle of triangle DEF)is the triangle with vertices at the midpoints of the triangle's sides DE, EF, and DF.	What I found helpful from your presentation is that not only you guided us but you explained why. I liked how you gave some definitions of some of the terms you were going to be using in your presentation but towards the end of your proof, I felt like you could've explain a bit more. That's the part I had a hard time understanding.
40	Luis	Medial Truangle	4	4	3	Triangle ABC with the vertices at the midpoints of the triangle's sides of AB, AC and BC.	The presentation was well detailed from the beginning to the end of the experiment. The presentation explained the step by step process of what was happening. The triangle that was created reminds me of something but I can't remember what of. The weakness of the presentation is that
41	Luis	The Medial Triangle	4	3	3	In any triangle ABC, we can create a medial triangle by connecting the midpoints of the sides of triangle ABC. Let the medial triangle be triangle DEF, so the altitudes of the medial triangle DEF are the same as the perpendicular bisectors of the opposite sides.	First of all, this worksheet has too many words. Luis knows his materials very well as we can see, but I think Luis should also concentrate on the last part of his worksheet more because he only showed one perpendicular as the altitude. Starting from the theorem, Luis should show us where the altitudes of the medial triangle are. Then, he can move on to the perpendicular bisectors of the created triangle. First, by the parallel lines, we will know that the altitudes are also the perpendiculars of the created triangle. Then, we measure the sides of the created triangle, and show that the vertices of the medial triangle are the midpoints of the created triangle. So, the altitudes of the medial triangle are the same as the perpendicular bisectors of the created triangle. Since this project does not require us to prove the theorem, so I suggest Luis spend more on the key points of the statement to make the idea clearer.
42	Luis	Medial Triangle	4	4	3	Any Triangle ABC has a unique triangle referred to as the medial triangle where its sides consist of sides on Triangle ABC and its corresponding angles are congruent.	I understand it is important to use words to describe what was going on, but here, it was just excessive. The text boxes, piled up on each other many times, and I could barely read what it said. There was a part of your theorem that stated that the triangle's vertices were midpoints, but I don't know if I'm really sold on that. It would have been great to see the values of each segment. Other than that, the variation of colors, the actual work and worksheet was great!
43	Luis	The medial Triangle	4	4	4	After drawing the the triangle EFD by drawing the parallel lines to the sides of the origin triangle ABC, The perpendicular bisectors of triangle ABC is the same as the altitudes of the medial triangle	Louis did an outstanding job, the fact that he give definition to the perpendicular bisector and the altitude on his open lab post helped a lot.
44	Luis	The Medial Triangle	3	3	3	The perpindicuar bisector of the side of the orginal triangle ABC are the aome as altitudes of the medial trangle DEF	I think it in very good idea that he created some qusetion theought out worksheet, I think that will make students understant the theorem better, but I also think that he did not clrarly shoe that that he is really trying to show in the worksheets , in otherwords I think he should have a concultion.
45	Majid	Ptolemy Theorem	3	2	3	A quadrilateral at a circle, the product of the diagonals is equal to the sum of the product of two pairs of opposite sides.	The theorem is easy to understand, but the worksheet does not show us how it works. For example, Majid used the measurement command to measure the sides and diagonals of the quadrilateral, but we cannot see if the result satisfies the equation of the theorem. The drag test will make no effect to the theorem without some more modifications. I think Majid can go back to the old assignment and find a way to write it effectively, so we including himself can actually see the theorem.
46	Majid	Ptolemy theorem	4	3	4	The Product of a quadrilaterals diagonals is equaled to the product of its bases plus a the product of its sides	I would have loved to see a ratio of comparison at the end. Truthfully, just looking at this, I would not believe the theorem. Watch the spelling at times also. Other then that, great work, very straight forward, very clean.
47	Majid	Ptolemy theorem	4	4	3	The product of the two diagonals is equal to the sum of the product of the two opposite segments.	The way you used different colors to distinguish the different segments was a very good idea. It made it easier to distinguish the segments. I also liked that you explained each step. The only weakness I see from your presentation is that you could've used the values of the segments into the theorem.
48	Majid	Ptolemy	4	4	4	There is a relationship between the four sides and the two diagonals of a quadrilateral.	The presentation was detailed and also shows step by step on how to recreate about what the theorem was stating. The weakness was that the size was small and that it was kinda hard to see some of the words.
49	Majid	Ptolemy Theorem	4	3	4	A circle with four points that form a quadrilateral has sides whose product of the opposite pairs and their sums is equal to the product of the two diagonals formed form the vertical angles of the quadrilateral.	What does center A have to do with the explanation of the Ptolemy Theorem, is there any relevance to the quadrilateral? Also, punctuation marks and Capital letters noting when your statements begin and end can be helpful in your explanation. But, besides that I understood the theorem you explained.
50	Majid	Ptolemy	4	3	3	Prove that the vertices of any two diagonal points between a four sided quadrilateral are equal to the sum of the remaining sides.	Would a drag test show that this holds true for any two diagonal points?
51	Majid	Ptolemy's Theorem	3	3	4	According to the expose, the Ptolemy Theorem is when the product of the diagonal equal to the sum of the product of the segment.	I think it could be better if he substitutes the letters by the actual values. Overall it was clear enough.
52	Mei	Simsons Theorem	3	3	4	There is a point p where perpendicular lines cross for all sides of some triangle ABC. When creating a point for each intersection of the perpendicular line and its respected triangle side, a line can be formed between all 3 intersections, thus creating the simson line.	Your geogebra file was good I understood it. At a certain slide I didn't understand it and I had to pause for a second to actually see what was going on. It was the slides where the green lines were formed.
53	Mei	Simpson's Theorem	4	4	4	The intersection of points of all 3 perpendicular and all 3 sidelines of a triangle inscribed in a circle from a point on a circle create a line called the Simpsons line.	I LOVE that you put in the step!!! it was so helpful!!! It made it so much easier!! Thank You.
54	Mei	simpson's theorem	3	4	4	After we create a triangle, we draw a circle that passes by this triangle vertices, then if we let P be a point in the circle such that it is different than the triangle vertices and we draw the three altitude that passes by P and perpendicular to the triangle sides, their points of the intersections with triangle sides lies on the same line.	Mei post about the Simpson on the open lab helped me understand the theorem, it also shows that she understand the theorem well and the instruction on the worksheet were really clear and precise.
55	Mei	Simson's Theorem	4	4	4	When moving point p throughout the circle the colinear points will always follow on the circle the same way the point moves.	The worksheet is detailed on the steps on how Mei created the triangle, colorful, labeled, and also the worksheet is interactive in which we are able to move the points on the worksheet to help us see what's going on.
56	Mei	Simpson's Line	4	4	4	If you have a triangle ABC, you can then make a circle using the vertices of the triangle. Then you pick a point anywhere on the on the circle label it and from that point you drop altitudes perpendicular to each line segment of the triangle. Then you connect two of the points that are connecting from the altitudes of your point made on the circle and then create a line through those points and you'll note a third point connected and this line is called Simpson's line.	I believe Mei's project was awesome, it took me a little while to figure out where Simpson's line came from, but I reviewed the video and I figured it out.
57	Mei	Simson's Line	4	3	3	Base on the project, I understand that the Simson's line is the line that creates the three intersections points that are collinear.	I believe that Mei could do a better job. She could clear it better such as cut off some of the lines and make sure the text is in separate section so that the presentation could be more effective.
58	Mei	Simson's Theorem	3	3	2	Simson's line will always contain the three points of intersection.	I think you're presentation would've been a bit more clearer if you have shown the text before performing it in geogebra.
59	Mei	Simsons Line	4	3	2	Given any triangle ABC you have a point P that lines on the circumcenter of the triangle. Use the perpendicular lines that lay on the sides of the triangle to show that the intersecting points on the altitudes always lie on the same line.	Step 2 I think for step 2 you should have included are we connecting two vertices a and b to c. and then connect ab together with another line. Then explain that you should have a triangle. Good use of color when you constructed the altitudes i can clearly depict them from the other lines. I think this is direct and straight to the point.
60	Sonam	The circumscribed Circle and the circumcenter	4	3	4	Each triangle has a unique circumscribed circle.	The steps on the worksheet were not in order, Step 2 showed up before the circle did, and I had no clue what was going on. Similarly, in the beginning, I could not tell what happened with the perpendicular lines because I could not see them. For some reason, I couldn't click them either. You did use the check boxes and slide shows effectively, and the color changes really helped me.
61	Sonam	The circumscribed circle and the circumcenter	3	2	2	A triangle has their own unique circumscribed circle with a center a point known as the circumcenter.	I think your presentation was overall easy to follow but I think it just needed to be more organized.
62	Sonam	The Circumscribed circle and the circumcenter	4	4	2	The outer circle formed from the vertices of a given triangle has a unique point called the circumcenter which is formed from the points made from the perpendicular bisectors of the sides of said triangle.	I had trouble with seeing your graph not sure if it was a problem on my end, but I believe a little more effort could of went into making it. Still, I understand the theorem you explained.
63	Sonam	The circumscribed Circle and the circumcenter	4	4	3	Every triangle ABC has a unique circumscribed circle.	I can see that in the presentation is detailed on how to explain the theorem and also the words show a step by step procedure. The weakness of the presentation is that the presentation picture of the triangle could have been a little more bigger because a lot of the words were hard the see.
64	Sonam	Circumscribed Circle	3	3	3	For any triangle ABC, we can create a circumscribed circle through the three vertices. The circumcenter of the triangle is the center of the created circle.	I think there might be some information missing from sheets 6-7 and 13-18. I would like to have some comments on how he stated the instructions. At first, he said that, "Construct the perpendicular bisector of one side of triangle." In fact, we are creating the perpendicular bisectors of each side of the triangle but not just for one side. I like how he used the checkbox to show how he got those points on the sides and the intersection of the perpendiculars. This is a good way to show that he knows what he is doing. For the angle measurement, I think Sonam can just measure the perpendicular angles. The angle measurement of the angles of the triangle is not important in this theorem. One more thing, there are some type errors in the statement of the theorem. I suggest writing as the following, every triangle has a unique circumscribed circle. The circumcenter of the triangle is the center of the circumscribed circle. I wish my comments were useful.
65	Sonam	circumcenter and circumscribed circle	3	3	3	In geometry,There is a unique circumscribed circle for each triangle and the both triangle and the circumscribed have the same circumcenter O.	Sonam eplained the theorem good on his open lab post and in the worksheet, the fact that he used different colors for each step will help someone new understand the theorem better.
66	Sonam	The Circumscribed Circle	3	3	3	prove that for any triangle ABC the perpendicular bisectors can be used to find the circumcenter of any given triangle.	I think if you spaced out the angles it would help the observer. Step one suggests only to find the perpendicular bisector of one side the wording should be changed so that students know each side. There are some mis-spelled words be careful next time just proof read everything.
67	Sonam	Simson's Theorem	3	4	4	With a Given triangle ABC and a point P on its circumcircle, if we draw 3 lines on Point P and perpindicular to line AB, BC and CA. Then thses 3 lins are collinear.	Her dynamic worksheets is very claen adn organized, but, if I am not wrong, it is better to use perpindicular line instead of altitudes in her sentence in #4
68	Tyniqua	Angle Bisector Concurrence Theorem	4	3	4	The bisectors of Triangle ABC has a point of concurrency where the point is equidistant from the sides of the circle.	It was a bit difficult to see where lines i,j,k came from but I then learned why, also, the angles came from out of nowhere. Great idea of the check box and questions on the sheet. The not was a bit confusing though. BTW, after reading this sentence you will know who's comment this was, but ITS THEOREM NOT THEORUM!! Great Job!!!
69	Tyniqua	Angle Bisector Theorem	2	2	3	The angle bisectors of a triangle are concurrent. The intersection of the angle bisectors of a triangle is called incenter, which is equidistant to the sides of of the triangle.	I think there are some information in the slides are missing: from slides 5-7, and from slides 9-11. The missing information makes it a little hard to understand the theorem. I like the circle which makes the property of equidistant stands out, but the actual work to show the incenter D is not clear. The little checkbox did not show us how she found the incenter. My suggestions will be following: first, create a triangle and creating the angle bisectors of the triangle. The intersection of the angle bisectors will be the incenter D. From the point D, we can make the altitudes to each side. Then, create a circle with the three feet of the altitudes. Then, we use the drag test, where we can see that the feet are always at the circle. In other words, the incenter D is equidistant to the sides of the triangle. At the same time, use some words to describe what she did will make her work understandable and nicer.
70	Tyniqua	Angle Bisector Concurrence Theorem	3	2	3	The incenter (point of concurrency) will always be in the interior of the triangle regardless how the angle bisectors extend.	I had a bit trouble understanding the theorem. I think you could have explained a bit more on each term just to make it easier to understand what was going on inside the triangle.
71	Tyniqua	Angle Bisector Concurrence	3	3	3	Any triangle ABC the interior angles are congruent.	The presentation is detailed later on in the slides and also explains what the theorem means and the theorem defined. The weakness of the presentation is that the slides should have said step by step on what was happening in the slides of the theorem.
72	Tyniqua	Angle Bisecting Concurrence	4	4	4	Prove that the angle bisectors of the interior angles of triangle ABC are concurrent.	Very organized! I think you did a fantastic job with the notes and question good use of pedagogy here.
73	Tyniqua	Angle bisctors theorem	3	3	3	Angle bisector concurrence theorem say that the inner center witch is the intersection of the angle bisectors on the triangle has the same distance from the midpoints of the sides, and this distance is a radius of a circle that lies inside the triangle.	Tyniqua did a good job but it would have been better if she add more information about the theorem in here open lab post.
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