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Butoxors [25]
3 years ago
5

Which congruence theorem can be used to prove △BDA ≅ △DBC? HL SAS AAS SSS

Mathematics
2 answers:
mario62 [17]3 years ago
8 0
HL, when you have 2 right triangles and their hypotenuses are congruent you are able to say HL
statuscvo [17]3 years ago
8 0

Answer:

A. Hypotenuse-leg (HL) congruence.

Step-by-step explanation:

We have been given a diagram of two right triangles and we are asked to determine the right congruence theorem that will prove △BDA ≅ △DBC.

Since we know that the hypotenuse-leg theorem states that if the hypotenuse and one leg of a right triangle are congruent to hypotenuse and corresponding leg of another right triangle, then the triangles are congruent.  

We can see from our diagram that hypotenuse(AB) of △BDA equals to hypotenuse (CD) of △DBC.  

We can see that triangles BDA and DBC share a common side DB.

Using Pythagorean theorem we will get,

CD^{2}=DB^{2}+BC^{2}...(1)  

AB^{2}=DB^{2}+AD^{2}...(2)

We have been given that CD=AB, Upon using this information we will get,

DB^{2}+BC^{2}=DB^{2}+AD^{2}

Upon subtracting DB^{2} from both sides of our equation we will get,

BC^{2}=AD^{2}

BC=AD

Therefore, by HL congruence △BDA ≅ △DBC.



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OB is the bisector of AD and EC. . EO = 2x – 20 and OC = 70 – x. Solve for x.
mixas84 [53]
<span>We know that OB bisects EC where they intersect would be at point O, 
so the two lengths EO and OC must be equal. 

2x - 20 = 70 - x 

</span>[adding x + 20 to each side] 

2x - 20 + x + 20 = 70 - x + x + 20

[simplifying]

<span>3x = 90 

</span>[dividing each side by 3]
<span>
x = 30 would be the answer</span>
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3 years ago
How do I solve this problem<br><br> Increased by 3.<br> 2 4 5 7 10 11 X, find the X
vitfil [10]

Answer:

36

Step-by-step explanation:

2+4+5+7+10+11=39

39-3

3 0
2 years ago
Guys its my birthday and ill be thankful if someone would help &lt; 3
MAXImum [283]

✔️ all real numbers

#VerifiedAnswer

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I WILL GIVE BRAINLIEST TO WHOEVER IS CORRECT
natali 33 [55]

Because M is the midpoint of AB, then AM and MB are equal distances. And because a segment can be written as the sum of its pieces, AM + MB = AB.

So,

AM + MB = AB                  <--- distance of a segment is the sum of its pieces

AM + AM = AB                  <--- M is the midpoint, so AM = MB

3x + 3 + 3x + 3 =  8x - 6   <--- substituting known amounts that were given

6x + 6 = 8x - 6                  <--- collect like terms on the left side

6x = 8x - 12                       <--- subtract 6 on both sides

-2x = -12                            <--- subtract 8x on both sides

x = 6                                  <--- divide both sides by -2

Because x = 6, we put it back into AM. 3(6) + 3 = 18 + 3 = 21


Thus, AM is 21 units.

3 0
3 years ago
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frozen [14]
Use a calculator and multiple
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