6. in this diagram, line segment cd is the perpendicular bisector of line segment ab. assume the conjecture that the set of points equidistant from a and b is the perpendicular bisector of ab is true. select all statements that must be true. a. ( am = bm ) b. ( cm = dm ) c. ( ea = em ) d. ( ea

Question

6. in this diagram, line segment cd is the perpendicular bisector of line segment ab. assume the conjecture that the set of points equidistant from a and b is the perpendicular bisector of ab is true. select all statements that must be true.
a. ( am = bm )
b. ( cm = dm )
c. ( ea = em )
d. ( ea < eb )
e. ( am < ab )
f. ( am > bm )
(from unit 1, lesson 3.)

7. the diagram was constructed with straightedge and compass tools. name all segments that have the same length as segment ( ac ).
(from unit 1, lesson 1.)

Accepted Answer

### Question 6 # Brief Explanations: - **Option A**: Since $ CD $ is the perpendicular bisector of $ AB $, by definition of a perpendicular bisector, it divides $ AB $ into two equal parts. So $ AM = BM $, this is true. - **Option B**: There is no information given that $ CD $ bisects itself at $ M $, so we cannot conclude $ CM = DM $, this is false. - **Option C**: Point $ E $ is not on the perpendicular bisector of $ AB $ (from the diagram), so we cannot say $ EA = EM $, this is false. - **Option D**: Since $ E $ is not on the perpendicular bisector of $ AB $, the set of points equidistant from $ A $ and $ B $ is the perpendicular bisector $ CD $. So $ E $ is not equidistant from $ A $ and $ B $, but actually, since $ M $ is on the bisector, $ EA$ and $ EB$: Wait, no, the conjecture says the set of points equidistant from $ A $ and $ B $ is the perpendicular bisector. So points not on $ CD $ are not equidistant. But in the diagram, $ E $ is not on $ CD $, so $ EA eq EB $. Wait, maybe I made a mistake. Wait, the perpendicular bisector is $ CD $, so any point on $ CD $ is equidistant from $ A $ and $ B $. $ E $ is not on $ CD $, so $ EA$ and $ EB $: Let's think again. The conjecture is "the set of points equidistant from $ A $ and $ B $ is the perpendicular bisector of $ AB $". So if a point is equidistant from $ A $ and $ B $, it's on $ CD $, and vice versa. So $ E $ is not on $ CD $, so $ EA eq EB $. But the option is $ EA < EB $? Wait, maybe I misread. Wait, no, let's check the diagram. $ M $ is the midpoint, $ CD \perp AB $. $ E $ is above $ CD $, not on $ CD $. So $ E $ is not equidistant from $ A $ and $ B $. But the problem says "select all statements that must be true". Wait, maybe I made a mistake with option D. Wait, no, let's re - evaluate: Wait, the conjecture is that the set of points equidistant from $ A $ and $ B $ is the perpendicular bisector ( $ CD $ ). So points on $ CD $ are equidistant from $ A $ and $ B $, points not on $ CD $ are not equidistant. But we don't know if $ EA < EB $ or $ EA>EB $ unless we have more info. Wait, maybe I messed up. Let's go back to option A: $ AM = BM $ because $ CD $ is the perpendicular bisector, so that's true. Option E: $ AM $ is half of $ AB $ (since $ M $ is the midpoint), so $ AM=\frac{AB}{2}$, so $ AM < AB $, this is true. Option F: $ AM = BM $, so $ AM>BM $ is false. Option B: No info about $ CM $ and $ DM $, false. Option C: $ EA = EM $? No, $ E $ is not related to $ M $ in that way. So the correct options are A, E? Wait, wait, maybe I made a mistake with option D. Wait, no, the conjecture is about equidistant points. So if $ E $ is not on $ CD $, then $ EA eq EB $, but we can't say $ EA < EB $ for sure. Wait, maybe the diagram shows that $ E $ is closer to $ A $? No, the diagram is not clear, but according to the definition: - $ CD $ is the perpendicular bisector of $ AB $, so $ AM = BM $ (A is true). - $ AM=\frac{AB}{2}$, so $ AM < AB $ (E is true). - There's no reason for $ CM = DM $ (B false), $ EA = EM $ (C false), $ EA < EB $ (we can't be sure, maybe D is false), $ AM>BM $ (F false). Wait, but maybe I was wrong about D. Wait, the conjecture is that the set of points equidistant from $ A $ and $ B $ is $ CD $. So $ E $ is not on $ CD $, so $ EA eq EB $. But is there a way to know $ EA < EB $? No, unless $ E $ is on the same side as $ A $, but the diagram doesn't show that. Wait, maybe the original problem has a different interpretation. Wait, let's check the options again: A. $ AM = BM $: True, because $ CD $ is the perpendicular bisector. B. $ CM = DM $: No, $ CD $ is a line, $ M $ is the intersection with $ AB $, no info about $ CM = DM $. C. $ EA = EM $: No, no relation given. D. $ EA < EB $: Since $ E $ is not on the perpendicular bisector, $ EA eq EB $, but we can't say $ EA < EB $ (maybe $ EA>EB $). E. $ AM < AB $: Since $ M $ is the midpoint, $ AM=\frac{AB}{2}$, so $ AM < AB $, true. F. $ AM>BM $: False, since $ AM = BM $. Wait, but maybe I made a mistake with D. Wait, the conjecture is that the set of points equidistant from $ A $ and $ B $ is $ CD $. So $ E $ is not on $ CD $, so $ EA eq EB $. But in the diagram, $ E $ is above $ CD $, and $ A $ and $ B $ are on either side of $ M $. Wait, maybe the problem considers that $ E $ is not on the bisector, so $ EA$ and $ EB $: Let's think of the perpendicular bisector theorem converse: If a point is equidistant from $ A $ and $ B $, it's on the perpendicular bisector. So the contrapositive: If a point is not on the perpendicular bisector, it's not equidistant from $ A $ and $ B $. So $ EA eq EB $, but we can't say $ EA < EB $ or $ EA>EB $ unless we have more info. So maybe D is false. So the correct options are A and E? Wait, but maybe the answer is A, E, and also maybe A, E, and another? Wait, let's check again. Wait, the problem says "line segment $ CD $ is the perpendicular bisector of line segment $ AB $". So by definition of a perpendicular bisector, it intersects $ AB $ at its midpoint and is perpendicular to $ AB $. So $ AM = BM $ (A is true), and $ AM=\frac{AB}{2}$, so $ AM < AB $ (E is true). For option B, there's no information that $ CD $ is bisected at $ M $, so $ CM $ and $ DM $ can be different. Option C: $ EA = EM $? No, $ E $ is a point not on $ CD $ or $ AB $ (from the diagram), so no. Option D: $ EA < EB $? We can't be sure. Option F: $ AM>BM $ is false. So the correct statements are A and E. ### Question 7 # Brief Explanations: The diagram is constructed with a straightedge and compass. In such constructions, when circles are drawn with the same radius, the radii of the same circle are equal, and if two circles intersect, the radii of each circle are equal. Looking at the diagram, the first circle has center $ A $ (or maybe $ C $ and $ A $ are related), wait, the circles: the first circle (left - most) has center at the mid - point? Wait, no, the diagram shows three circles. The first circle (with center, let's say, the center of the left - most circle is such that $ AC $ is a radius. Then the middle circle has center at $ A $ (or $ B $)? Wait, actually, when constructing with compass, if we draw a circle with center $ A $ and radius $ AC $, and another circle with center $ B $ and radius equal to $ AC $, and a third circle with center $ B $ and radius equal to $ AC $ (or $ AB $)? Wait, no, looking at the line $ CD $ with points $ C, A, B, E, D $. The left - most circle has center at the center of the first circle, with radius $ AC $. The middle circle has center at $ A $ (or $ B $) with radius $ AB $ (but $ AC = AB $ because of the compass construction). Wait, actually, in compass constructions, when you draw a circle with center $ A $ and radius $ AC $, and then a circle with center $ B $ with the same radius, and then another circle with center $ B $ with the same radius. So the segments equal to $ AC $ are $ AD $? No, wait, let's see: The left - most circle: center is, let's say, the center of the first circle, and $ AC $ is a radius. Then the middle circle (center $ A $) has radius $ AB $, but if $ AC = AB $ (because the compass was used with the same radius), then $ AB = AC $. Then the right - most circle (center $ B $) has radius $ BE $ or $ BD $? Wait, no, the correct approach is: In the diagram, the circles are constructed such that $ AC = AB = BE = BD $. Wait, let's think again. When you draw a circle with center $ A $ and radius $ AC $, and then a circle with center $ B $ with the same radius (so $ AB $ is equal to the radius, so $ AC = AB $). Then, if we draw another circle with center $ B $ with the same radius, then $ BE = AB = AC $, and $ BD = AB = AC $. Wait, no, the line is $ C---A---B---E---D $. So the length $ AC $ is equal to $ AB $ (because the first two circles are drawn with the same compass radius), $ AB $ is equal to $ BE $ (because the middle and right - most circles are drawn with the same compass radius), and $ BE $ is equal to $ BD $ (because the right - most circle has radius $ BE $ and $ BD $ is a radius of the right - most circle). Wait, no, the right - most circle has center at $ B $ (or maybe the center of the right - most circle is such that $ BD $ is a radius). Wait, actually, the correct segments equal to $ AC $ are $ AB $, $ BE $, and $ BD $. Wait, let's check: - $ AC $: radius of the left - most circle. - $ AB $: radius of the middle circle (since the middle circle is drawn with center $ A $ and radius equal to $ AC $, so $ AB = AC $). - $ BE $: radius of the right - most circle (drawn with center $ B $ and radius equal to $ AB = AC $, so $ BE = AB = AC $). - $ BD $: radius of the right - most circle (since $ BD $ is a radius of the right - most circle, and the radius is equal to $ AC $, so $ BD = AC $). So the segments with the same length as $ AC $ are $ AB $, $ BE $, and $ BD $. ### Question 6 Answer: A. $ AM = BM $, E. $ AM < AB $ ### Question 7 Answer: The segments with the same length as $ AC $ are $ AB $, $ BE $, and $ BD $

Sovi.AI - AI Math Tutor

QUESTION IMAGE

Question

Explanation:

Question 6

Question 6 Answer:

Question 7 Answer:

Answer:

Question 6 Answer:

Question 7 Answer: