Can some please explain how to do #37 to me.... i'm not really sure how. http://prntscr.com/8dqn9t
@pooja195, do you remember the acronym for congruent triangles?
No :/
You don't remember what CPCTC stands for?
Corresponding parts of a triangle are congruent...? right?
Correct, Corresponding parts of congruent triangles are congruent. What do you think this means for \(\triangle{ABC} \cong \triangle{DEF}\) ?
that equal sign with the squiggly thingy means that they correspond so triangle ABC corresponds to triangle DEF?
it means congruent or equal trinagles
same thing -.-
Well, @pooja195, here's a bit of insight for you: \(\triangle{ABC} \cong \triangle{DEF}\) means the following: \(\angle{A} \cong \angle{D}\) \(\angle{B} \cong \angle{E}\) \(\angle{C} \cong \angle{F}\) \(\overline{AB} \cong \overline{DE}\) \(\overline{BC} \cong \overline{EF}\) \(\overline{AC} \cong \overline{DF}\)
this is confusing... :/
The corresponding angles have the same measure and the corresponding segments have the same length.
So DEF is 5?
I think what you meant to say is that since \(\overline{AC} = 5\) then \(\overline{DF} = 5\)
@pooja195, what does the problem ask us to find?
Find DF and m<B
Okay, and we found \(\overline{DF}\). Now we have to find \(m\angle{B}\). Any ideas on that at all based on what was explained above?
B and E are congruent
B and E correspond with each other and therefore have the same measure.
so 100 degrees?
\(m\angle{E} = 100^{\circ}\), therefore \(m\angle{B} = 100^{\circ}\). That's how you would say it properly. Saying just "100 degrees" isn't really saying anything other than just "announcing" 100 degrees. Somebody might think you're referring to the temperature outside.
Aight thanks :)
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