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)
10/17= 0.588
So it's 59%
This is correct. :)
Answer:
B. two
explanation:
i took the test
hope this helps. have a great day!!
Answer:
2
Step-by-step explanation:
1,3,4 are arithmetic sequences
2 is a geometric sequence
<span>the state or fact of being similar basically saying the same
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