In the figure above, ΔABC is a scalene triangle, DE = BC. Using D, and E as two vertices of a new triangle DEF, so that ΔDEF ≅ΔBCA, how many triangles can be drawn?
#Is the answer just 1?
stuck birch
stray jettyBOT
stuck birch
<@&286206848099549185>
woeful chasm
A can be above or below
stuck birch
so 2?
tall radish
You can have it lean to the left or right
Is it still congruent in that case actually
tall radish
The triangle thats drawn the point A is more to the right
You can imagine reflecting it over a vertical line
Whats the definition of congruence actually
That you can reach it using some transformation?
stuck birch
congruency means all same angle and same sides
tall radish
How many guesses do you have
I would try like 4 next but idk i would want to think more about it
tall radish
I can draw my guess
stuck birch
the answer choices are 1, 2, 3, 4, or 6
tall radish
this is what i'm thinking currently
that's what I meant by left or right
but maybe there's more
ok so since DE = BC that actually restricts things
you only have two choices of lengths to draw the other sides with
what do you think
stuck birch
i think it probably should be 4 then
because i have 2 choices AB > AC or AB < AC. and both times i can reflect over
tall radish
yeah exactly 2*2=4
but can we get 6 somehow?
i'm thinking no because that DE is fixed
and it's obvious none of the two triangles are the same (if two were the same, it would make it 3)... since it's scalene?
so it should be 4
i can't think of how to make 3 or 6
stuck birch
wait but based on this logic couldn't there be infinitely many because we can move point A anywhere and match it with point F
tall radish
wait wdym
so you have a triangle ABC that's scalene meaning that all the side lengths are different, also the triangle is like fixed before you start considering things
if the triangle DEF is congruent to ABC and you've also fixed the segment DE somewhere, you can't draw a third point arbitrarily, or else you might get a triangle with different side lengths?
maybe i'm not understanding what you mean
stuck birch
ok so what I am saying is that, we can make BCA have BA = 4 and CA = 3. We can also make triangle DEF so DF = 4 and EF = 3. We have basically infinitely ways to make the 2 triangles congruent no?
maybe I am not understanding the question correctly
tall radish
why would there be infinitely many ways
if we imagine you have like two sticks of length 4 and 3, and say the third side of the triangle is the ground right
you can't move them around arbitrarily and still have a triangle
like only at certain points will the two sticks perfectly touch each other and form a triangle?
stuck birch
maybe I am not understanding the question properly
is the question asking saying that triangle BCA has a fixed length in which how many ways that you can make triangle DEF congruent to BCA?
tall radish
I think it's saying like, you are starting with a scalene triangle ABC. We give you a side DE = BC. How many ways can you draw the point F in the plane so that DEF is congruent to ABC?
that's how I'm understanding it
stuck birch
hmmm i see
so the answer probably is just 4 then right
because it is basically asking how many ways you can reflect triangle BCA
tall radish
yea but don't hold me accountable if it's not right lol
tall radish
because it is basically asking how many ways you can reflect triangle BCA
yeah it's asking how many ways you can create the same triangle given a fixed side
which I think limits it to reflection
and you can only do vertical and horizontal reflections here i think
stuck birch
yes I inputted 4 in and it is correct
tall radish
ok nice
stuck birch
thanks so much
tall radish
np
echo hill
.solved