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Mathematics 17 Online
OpenStudy (anonymous):

can someone help

OpenStudy (anonymous):

OpenStudy (anonymous):

i got triangle BCX congruent to triangle BAX by SSS but idk where to go from there

OpenStudy (oaktree):

You have that angle CXA and angle FXD are congruent by opposite angles, and since in a parallelogram opposite angles are congruent, angle B is congruent to angle CXA and angle E is congruent to angle FXD. Get it?

OpenStudy (anonymous):

no because fxde isnt a paralellogram

OpenStudy (anonymous):

so how do you do the opp angles

OpenStudy (oaktree):

Yeah it is. You have it given to you.

OpenStudy (oaktree):

Look at the question. It say that both are parallelograms.

OpenStudy (anonymous):

OH sorry i didnt see that

OpenStudy (oaktree):

Yeah, that would help. XD

OpenStudy (oaktree):

So now do you understand the proof?

OpenStudy (anonymous):

yeah but like i dont understand the steps its like 2 steps

OpenStudy (anonymous):

wait also in general can you use opp <'s are congruent and opp sides are congruent in the same proof/same parallellogram or do you only do one?

OpenStudy (anonymous):

@jim_thompson5910

jimthompson5910 (jim_thompson5910):

ABCX is a parallelogram (given) so the opposite angles are congruent. You can prove this by dividing up the parallelogram into two triangles. Use the properties of parallel lines to find a pair of congruent alternate interior angles. Then you can combine that with the diagonal to use the ASA property. Finally, use CPCTC.

jimthompson5910 (jim_thompson5910):

So because the opposite angles are congruent, we know that < B = < X since B is opposite X

jimthompson5910 (jim_thompson5910):

similarly, DXFE is a parallelogram, so < E = < X because X is opposite E

OpenStudy (anonymous):

yeah i understand all that but after i prove the two triangles in parallelogram CBAX where do i go from there?

jimthompson5910 (jim_thompson5910):

by the transitive law, we know that if < B = < X and < X = < E, then < B = < E

OpenStudy (anonymous):

so do i do the same thing in parallelogram FXDE as i did in BAXC?

OpenStudy (anonymous):

and then what is the statement for the CPCTC?

OpenStudy (anonymous):

@jim_thompson5910

jimthompson5910 (jim_thompson5910):

CPCTC stands for Corresponding Parts of Congruent Triangles are Congruent

OpenStudy (anonymous):

yeah but whats the statement to go along with that

jimthompson5910 (jim_thompson5910):

in a nutshell it means that if you have 2 congruent figures, then the corresponding parts (ie the parts that pair up) will also be congruent

jimthompson5910 (jim_thompson5910):

oh, the opposite angles being congruent

OpenStudy (anonymous):

so <B and <x

OpenStudy (anonymous):

i mean <B and <E

OpenStudy (anonymous):

but i thought the reason they were congruent was because of transitive

OpenStudy (anonymous):

@jim_thompson5910

jimthompson5910 (jim_thompson5910):

the opposite angles are congruent because of CPCTC you'll have < B = < X and < X = < E based on the property

jimthompson5910 (jim_thompson5910):

once you have < B = < X and < X = < E, you use the transitive property to say < B = < E

OpenStudy (anonymous):

oh okay thanks! so just to sum it up: make the 2 parallelograms into 2 triangles each, make the parallelograms congruent, cpctc, transitive

jimthompson5910 (jim_thompson5910):

prove the triangles congruent (not the parallelograms), but yes you have the basic outline

OpenStudy (anonymous):

kk thanks so much! also can i use more than one method in one parallelogram? like opp sides are cong. and opp angles are cong.

jimthompson5910 (jim_thompson5910):

yeah if you've proven a property, you don't need to go through the same work to prove a very similar thing (just with different labels)

jimthompson5910 (jim_thompson5910):

so once you show that opposite angles in a parallelogram are congruent, you can just refer to that idea (instead of reinventing the wheel each time)

OpenStudy (anonymous):

ok thanks!!!

jimthompson5910 (jim_thompson5910):

sure thing

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