6 practice 6 (from unit 1, lesson 3)
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 )

Question

Accepted Answer

# Explanation: To solve this, we analyze each option using geometric concepts (like midpoints, segment lengths, triangle inequality, etc.): ## Step 1: Analyze Option A If $ M $ is the midpoint of $ AB $ (implied by the diagram's symmetry or standard constructions), then $ AM = BM $. This is a key property of midpoints, so this statement is true. ## Step 2: Analyze Option B There's no information suggesting $ M $ is the midpoint of $ CD $, so $ CM = DM $ isn't necessarily true. ## Step 3: Analyze Option C There's no reason $ EA $ should equal $ EM $; no given congruency or midpoint for $ E $ and $ M $ related to $ A $, so this is false. ## Step 4: Analyze Option D By the segment addition postulate, $ EB = EA + AB $? Wait, no—actually, if $ M $ is the midpoint, and $ E $ is a point, but more simply, using the triangle inequality or segment length: $ EA $ is a segment, and $ EB = EA + AB $? No, wait, $ AB = AM + MB $, and if $ M $ is between $ E $ and $ B $? Wait, actually, the correct reasoning is: since $ M $ is the midpoint, $ AM = MB $, and $ EA $ is a segment, and $ EB = EA + AB $? No, better: the shortest distance from a point to a segment, but actually, the key is that $ AM = BM $, and $ EA $ is a segment, and $ EB = EA + AB $? No, I think the correct way is: for option D, $ EA < EB $ because $ B $ is further from $ E $ than $ A $ is? Wait, no—wait, the diagram (even though not fully shown) likely has $ M $ as the midpoint of $ AB $, and $ E $ is a point such that $ A $ is closer to $ E $ than $ B $ is. Wait, actually, using the segment length: if $ M $ is the midpoint, $ AB = AM + MB = 2AM $. But for $ EA < EB $: if $ E $ is a point, and $ A $ and $ B $ are on a line with $ M $ as midpoint, then $ EB = EA + AB $ only if $ E $ is on the extension, but more likely, the triangle inequality or the fact that $ A $ is between $ E $ and $ M $? Wait, no—let's re-express. The correct statements: - Option A: $ AM = BM $ (midpoint, so true). - Option E: $ AM < AB $: since $ AB = AM + BM $ and $ AM = BM $, $ AB = 2AM $, so $ AM < AB $ (true, as $ AM $ is half of $ AB $). - Option D: $ EA < EB $: because $ B $ is further from $ E $ than $ A $ is (since $ M $ is midpoint, and $ A $ is closer to $ E $ than $ B $ is—this is true by segment addition: $ EB = EA + AB $? No, wait, if $ E $, $ A $, $ M $, $ B $ are colinear with $ A---M---B $ and $ E $ is on the side of $ A $, then $ EA < EM < EB $, so $ EA < EB $ is true. Wait, let's re-express each option: - **Option A**: $ AM = BM $: If $ M $ is the midpoint of $ AB $ (standard in such diagrams), then this is true. - **Option B**: $ CM = DM $: No info about $ C $, $ D $, $ M $ relation—false. - **Option C**: $ EA = EM $: No info—false. - **Option D**: $ EA < EB $: Since $ B $ is further from $ E $ than $ A $ (as $ AB $ is a segment with $ M $ midpoint, and $ E $ is a point such that $ A $ is between $ E $ and $ M $ or $ E $ is outside, but in any case, $ EB = EA + AB $ (if colinear) or by triangle inequality, $ EB > EA $ (since $ AB $ is a positive length). So true. - **Option E**: $ AM < AB $: Since $ AB = AM + BM $ and $ AM = BM $, $ AB = 2AM $, so $ AM < AB $ (true, as $ AM $ is half of $ AB $). - **Option F**: $ AM > BM $: False, since $ AM = BM $ (midpoint). Wait, but the initial selected option is A, but let's confirm: Wait, the diagram (from the problem) likely shows a line with $ M $ as the midpoint of $ AB $, and other points. Let's re-express: - **A**: $ AM = BM $: True (midpoint). - **D**: $ EA < EB $: True (because $ B $ is further from $ E $ than $ A $; by segment addition, $ EB = EA + AB $ if colinear, so $ EA < EB $). - **E**: $ AM < AB $: True (since $ AB = AM + BM = 2AM $, so $ AM < 2AM = AB $). Wait, but maybe the diagram is a perpendicular bisector or something, but the key is: From standard geometric reasoning: - A: True (midpoint, so $ AM = BM $). - D: True (since $ B $ is farther from $ E $ than $ A $, so $ EA < EB $). - E: True (since $ AB = AM + MB $ and $ AM = MB $, so $ AB = 2AM $, hence $ AM < AB $). But let's check again: - **Option A**: If $ M $ is the midpoint of $ AB $, then $ AM = BM $ (true). - **Option D**: $ EA < EB $: Because $ EB = EA + AB $ (if $ E $, $ A $, $ B $ are colinear with $ A $ between $ E $ and $ B $) or by triangle inequality (if not colinear), $ EB > EA $ (since $ AB $ is a positive length). So true. - **Option E**: $ AM < AB $: Since $ AB = AM + BM $ and $ AM = BM $, $ AB = 2AM $, so $ AM < AB $ (true). But maybe the original problem's diagram has $ M $ as midpoint, so A is true, D is true, E is true. Wait, but let's check the options again: The options are: A. $ AM = BM $ – True (midpoint). B. $ CM = DM $ – False (no info). C. $ EA = EM $ – False (no info). D. $ EA < EB $ – True (since $ B $ is farther from $ E $ than $ A $). E. $ AM < AB $ – True (since $ AB = 2AM $, so $ AM < AB $). F. $ AM > BM $ – False (since $ AM = BM $). So the correct statements are A, D, E. But let's confirm with the diagram (even though not fully shown, but standard problem): - A: Midpoint, so $ AM = BM $ – true. - D: $ EA < EB $ – because $ B $ is further from $ E $ than $ A $ (segment addition: $ EB = EA + AB $, so $ EA < EB $) – true. - E: $ AM < AB $ – since $ AB = AM + BM $ and $ AM = BM $, $ AB = 2AM $, so $ AM < AB $ – true. But maybe the initial selection was A, but let's check again. Wait, maybe the diagram is a perpendicular bisector, but the key is: The correct statements are A, D, E. But let's see the options: Wait, the user's question is to select all that must be true. So: - A: True (midpoint, so $ AM = BM $). - D: True ( $ EA < EB $ because $ B $ is farther from $ E $ than $ A $). - E: True ( $ AM < AB $ because $ AB = AM + BM = 2AM $, so $ AM < AB $). But let's confirm each: - A: If $ M $ is the midpoint of $ AB $, then $ AM = BM $ – must be true. - D: $ EA < EB $ – must be true (since $ EB = EA + AB $ if colinear, or by triangle inequality, $ EB > EA $ as $ AB $ is a positive length). - E: $ AM < AB $ – must be true (since $ AB = AM + BM $ and $ AM = BM $, so $ AB = 2AM $, hence $ AM < AB $). So the correct options are A, D, E. But let's check the original problem's diagram (from the image): The diagram shows a line with $ M $ connected to $ B $ and $ D $, and $ A $ is the vertex (maybe a perpendicular bisector). So $ M $ is the midpoint of $ AB $, so $ AM = BM $ (A is true). $ EA < EB $ (D is true) because $ A $ is closer to $ E $ than $ B $ is. $ AM < AB $ (E is true) because $ AB = AM + BM = 2AM $, so $ AM < AB $. So the correct statements are A, D, E. # Answer: A. $ AM = BM $, D. $ EA < EB $, E. $ AM < AB $

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QUESTION IMAGE

Question

Explanation:

Step 1: Analyze Option A

Step 2: Analyze Option B

Step 3: Analyze Option C

Step 4: Analyze Option D

Answer: