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Llana [10]
3 years ago
9

Please help me withe this proof as well.​

Mathematics
1 answer:
baherus [9]3 years ago
4 0

- D is the midpoint of AB, E is the midpoint of BC

Answer: A. Given

I left off DB||FC because that's not given.  But we can construct it.

Construct line through C parallel to AB.  Extend DE to intersect and call the meet F.

- DB || FC

By Construction

----

- Angle B congruent to angle FCE

Answer: D. Alternate Interior Angles

We have transversal BC across parallel lines AB and CF, so we get congruent angles ABC and FCB aka FCE

- angle BED congruent to angle CEF

Answer: H. Vertical angles are congruent

When we get lines meeting like this we get the usual congruent and supplementary angles.

- Triangle BED congruent to Triangle CEF

Answer: F. Angle Side Angle

We have BE=CE, DBE=FCE, BED=CEF

- DE congruent to FE and DB congruent to FC

Answer: C. CPTCTF

Corresponding parts ...

- AD congruent to DB and DB congruent to FC therefore AD congruent to FC

Answer: E. Transitive Property of Congruent

Things congruent to the same thing are congruent

- ADFC is a parallelogram

Answer: G.  AD and FC are congruent and parallel

Presumably this is a theorem we have already established.

- DE || AD

Answer: B. Definition of a parallelogram

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Answer:

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Step-by-step explanation:

We can create a triangle from the information given.

<u>We can call the starting point (0,0) in the coordinate system.</u>

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Solving for x:

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3 years ago
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The distance between any 2 points P(a,b) and Q(c,d) in the coordinate plane, is given by the formula:<span>

<span> |PQ|= \sqrt{ (a-c)^{2} + (b-d)^{2}}</span></span>


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3 years ago
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I think that is the correct answer sorry I didn't answer before didn't see the problem.

7 0
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Answer:

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