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liubo4ka [24]
3 years ago
8

In the adjoining equilateral triangle PQR , X , Y and Z are the middle points of the sides PQ , QR and RP respectively. Prove th

at XYZ is also an equilateral triangle.
~Thanks in advance ! ​

Mathematics
2 answers:
chubhunter [2.5K]3 years ago
7 0

Answer:

this is your answer. thanks, for great point

olasank [31]3 years ago
5 0

Answer:

See Below

Step-by-step explanation:

<u>Statements:</u>                                                        <u>Reasons:</u>

<u />1)\text{ } \Delta PQR\text{ is equilateral}                                        Given

2)\text{ }PQ=QR=RP                                               Definition of Equilateral

3)\text{ } X, Y, Z\text{ are the midpoints of } PQ, QR, RP       Given

4)\text{ } PX=XQ                                                        Definition of Midpoint

5)PQ=PX+XQ                                               Segment Addition

6)\text{ } PQ=2XQ                                                      Substitution

7)\text{ } QY=YR                                                        Definition of Midpoint

8)\text{ }QR=QY+YR                                               Segment Addition

9)\text{ } QR=2QY                                                       Substitution

10)\text{ } 2XQ=2QY                                                  Substitution

11)\text{ } XQ=QY                                                      Division Property of Equality

12)\text{ }  XQ=PX=QY=YR                                Transitive Property

13)\text{ } RZ=ZP                                                       Definition of Midpoint

14)\text{ } RP=RZ+ZP                                             Segment Addition

15)\text{ } RP=2RZ                                                     Substitution

16)\text{ } 2XQ=2RZ                                                   Substitution

17)\text{ } XQ=RZ                                                       Substitution

18)\text{ } XQ=PX=QY=YR=RZ=ZP            Transitive Property

19)\text{ } \angle Q\cong \angle R\cong \angle P                                             Definition of Equilateral

20)\text{ } \Delta XQY\cong \Delta YRQ \cong \Delta ZPX                          SAS Congruence

21)\text{ } XY\cong YZ\cong ZX                                           CPCTC

22)\text{ } \Delta XYZ\text{ is equilateral}                                     Equilateral Triangle Theorem

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y= 3\cos(2(x + \pi/2)) - 2

The options are:

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y= 3\sin(2(x + \pi/4)) - 2\\y= 3\sin (2x + \pi/2) -2\\y = -3\cos(2x) -2\\y = 3\cos(2x + \pi) -2\\y = 3\cos(2(x+ \pi/2)) -2

(the last second step was the use of the fact that cos flips its sign after pi radian increment in its input)
Thus, this option is same as the given function.

  • Option 2: y= -3\sin(2(x + \pi/4)) - 2

This option if would be true, then from option 1 and this option, we'd get:
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Thus, this option is not same as the given function.

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