can someone help
i got triangle BCX congruent to triangle BAX by SSS but idk where to go from there
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?
no because fxde isnt a paralellogram
so how do you do the opp angles
Yeah it is. You have it given to you.
Look at the question. It say that both are parallelograms.
OH sorry i didnt see that
Yeah, that would help. XD
So now do you understand the proof?
yeah but like i dont understand the steps its like 2 steps
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?
@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.
So because the opposite angles are congruent, we know that < B = < X since B is opposite X
similarly, DXFE is a parallelogram, so < E = < X because X is opposite E
yeah i understand all that but after i prove the two triangles in parallelogram CBAX where do i go from there?
by the transitive law, we know that if < B = < X and < X = < E, then < B = < E
so do i do the same thing in parallelogram FXDE as i did in BAXC?
and then what is the statement for the CPCTC?
@jim_thompson5910
CPCTC stands for Corresponding Parts of Congruent Triangles are Congruent
yeah but whats the statement to go along with that
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
oh, the opposite angles being congruent
so <B and <x
i mean <B and <E
but i thought the reason they were congruent was because of transitive
@jim_thompson5910
the opposite angles are congruent because of CPCTC you'll have < B = < X and < X = < E based on the property
once you have < B = < X and < X = < E, you use the transitive property to say < B = < E
oh okay thanks! so just to sum it up: make the 2 parallelograms into 2 triangles each, make the parallelograms congruent, cpctc, transitive
prove the triangles congruent (not the parallelograms), but yes you have the basic outline
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.
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)
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)
ok thanks!!!
sure thing
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