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KIM [24]
4 years ago
13

Given: Base ∠BAC and ∠ACB are congruent.

Mathematics
2 answers:
Sergio039 [100]4 years ago
8 0

Refer to the image attached.

Given: \angle BAC and \angle ACB are congruent.

To Prove: \DeltaABC is an isosceles triangle.

Construction: Construct a perpendicular bisector from point B to Line segment AC.

Consider triangle ABD and BDC,

\angle BAD = \angle BCD (given)

\angle ADB = \angle BDC = 90^\circ

(By the definition of a perpendicular bisector)

AD=DC (By the definition of a perpendicular bisector)

Therefore, \Delta ABD \cong \Delta BDC by Angle Side Angle(ASA) Postulate.

Line segment AB is congruent to Line segment BC because corresponding parts of congruent triangles are congruent.(CPCTC)

BartSMP [9]4 years ago
6 0
Blank 1: *ASA Postulate*
Blank 2: *CPCTC*

Based on the info. your given and I drew (screenshot). Blank 1 is ASA Post. and Blank 2 is CPCTC because of ASA Post.

i am so sorry!
I meant to draw line segment AD is congruent to AC.... However Line segment AB is congruent to BC....

You might be interested in
Find all real numbers t such that (2/3)t - 1 < t + 7 ≤ -2t + 15. Give your answer as an interval.
Andrews [41]
Break up chain with AND statement:

(2/3)t - 1 < t + 7 AND <span>t + 7 ≤ -2t + 15

left side, we solve for t
</span>
<span>(2/3)t - 1 < t + 7
(2/3)t - t < 7 + 1
(-1/3)t < 8
t > -24

right side, we solve for t</span>
<span>t + 7 ≤ -2t + 15
t + 2t </span><span>≤ 15 - 7
3t </span><span>≤ 8
t </span><span>≤ 8/3

So our answer would be </span>
t > -24 AND t <span>≤ 8/3

as an interval, this is
</span>-24 < t <span>≤ 8/3
or in interval notation
(-24, 8/3]</span>
8 0
4 years ago
Who will solve this attached picture problem correctly they will be marked as Brainliest answer.​
Elodia [21]

Answer:

no it isn't

Step-by-step explanation:

if you find all of the angles using the angle addition postulate and supplementary angles rule (add to 180 degrees), you will be left with an angle that is not able to satisfy either of these rules, 75\neq60

3 0
3 years ago
Five years back, Michael's age was 2/3 the age of James. After ten years he will be 5/6 of the age of James. What are the equati
irakobra [83]
I find it easier to start from scratch and write out the equations, then compare them with the given equations.

Let m and j represent the ages of the two boys.  Then m-5=(2/3)(j-5)
Also, in ten years:  m+10 = (5/6)j.

Let's solve this system:  Mult the first eqn by 3 to remove the fraction:
3(m-5) = 2(j-5), or 3m - 15 = 2j -10.  Next, mult. the 2nd eqn by 6 to remove the fraction:  6m+60 = 5j.

Our two equations are (at this point)      3 m - 15 = 2j - 10 and
                                                                6m  +60  = 5j

Let's mult. the first equation by -2, so as to obtain the coefficient -6 for m:

-6m + 30 = -4j + 10
 6m + 60  = 5j
---------------------------
90 = j + 10, so j = 80 (wow!)

Let's now find m:  6m + 60 = 5(80), or    6m = 400+60 = 460

Then m = 76 2/3.

We must check our solution (m = 76 2/3, j = 80):

Let's subst. these values into     m-5=(2/3)(j-5)

Does 76 2/3 - 5 = (2/3)(80) - 10/3?

Does 76 2/3 - 15/3 = 160/3 - 10/3?

Does 76 - 15 = 160 - 10    No.  So something's wrong here.

Going back to the problem statement:

<span>x-5=2/3(y-5) and x+10=5/6(y+10)  seems correct!


Mike's age 5 years ago was x-5, and james' age 5 years ago was y-5.  The multiplier (2/3) is also correct.

Mikes age 10 years from now will be x+10, and james' y+10.

The first answer set is the correct one.</span>
7 0
3 years ago
Plssssss helppppppppppp​
lubasha [3.4K]
I believe the answer is A
5 0
3 years ago
Read 2 more answers
Giving brainliest!!!!!!!!!!
son4ous [18]
It’s A i did the test myself..
5 0
3 years ago
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