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viva [34]
3 years ago
15

Read the proof. Given: AB ∥ DE Prove: △ACB ~ △DCE We are given AB ∥ DE. Because the lines are parallel and segment CB crosses bo

th lines, we can consider segment CB a transversal of the parallel lines. Angles CED and CBA are corresponding angles of transversal CB and are therefore congruent, so ∠CED ≅ ∠CBA. We can state ∠C ≅ ∠C using the reflexive property. Therefore, △ACB ~ △DCE by the AA similarity theorem. SSS similarity theorem. AAS similarity theorem. ASA similarity theorem.

Mathematics
2 answers:
White raven [17]3 years ago
8 0
Pls. see attachment. 

pogonyaev3 years ago
7 0

Answer:

(A) AA similarity

Step-by-step explanation:

Given: AB is parallel to DE.

To prove: △ACB is similar to △DCE.

Proof: It is given that AB is parallel to DE, thus because the lines are parallel and segment CB crosses both lines, we can consider segment CB a transversal of the parallel lines.

Now, from △ACB and △DCE, we have

∠CED≅∠CBA (corresponding angles of transversal CB and are therefore congruent)

and ∠C ≅ ∠C (Reflexive property)

Thus, by AA similarity postulate,

△ACB is similar to △DCE

Hence proved.

Thus, option A is correct.

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If a+b+c =0 show that a³+b³+c³= 3abc
slega [8]

Answer:

Step-by-step explanation:

a+b+c=0, a+b=-c,a+c=-b, b+c=-a

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a^3+ab^2+ac^2+2a^2b+2a^2c+2abc+a^2b+b^3+bc^2+2ab^2+2abc+2b^2c+a^2c+b^2c+c^3+2abc+2ac^2+2bc^2=a^3+b^3+c^3+3a^2b+3a^2c+3ac^2+3ab^2+3bc^2+3b^2c+6abc=

a^3+b^3+c^3+3a^2*(b+c)+3c^2(a+b)+3b^2(a+c)+6abc=

a^3+b^3+c^3+3a^2*(-a)+3c^2*(-c)+3b^2*(-b)+6abc=

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6abc-2a^3-2b^3-2c^3=2(3abc-a^3-b^3-c^3)=

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7 0
3 years ago
If f(x)=x2+10sinx, show that there is a number c such that f(c)=1000.
gayaneshka [121]
F(x) is continuous for all x.

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slamgirl [31]

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and slope m = \frac{(-3-3)}{0-(-2)}

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6 0
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enot [183]
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