Given:ABCD is a rhombus.
To prove:DE congruent to BE.
In rombus, we know opposite angle are equal.
so, angle DCB = angle BAD
SINCE, ANGLE DCB= BAD
SO, In triangle DCA
angle DCA=angle DAC
similarly, In triangle ABC
angle BAC=angle BCA
since angle BCD=angle BAD
Therefore, angle DAC =angle CAB
so, opposite sides of equal angle are always equal.
so,sides DC=BC
Now, In triangle DEC and in triangle BEC
1. .DC=BC (from above)............(S)
2ANGLE CED=ANGLE CEB (DC=BC)....(A)
3.CE=CE (common sides)(S)
Therefore,DE is congruent to BE (from S.A.S axiom)
Answer:
it y = 2x + b
Step-by-step explanation:
YOU NEED TO PUT THE M FIRST NOT THE X
Answer:
A rotation and a translation
or
rotation
Step-by-step explanation:
Again, we have to rotate to make the shape fit in, right?
But it wouldn't fit... Or will it? If you can take 2 guesses, you can try the rotation and a translation or the rotation.
Sorry, I'm also getting confused at some point :(!
Assuming that the rectangle is split evenly, we see three portions, two of which are shaded.
100% / 3 = 33.3% * 2 = 66.6% shaded
