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:4.18
Step-by-step explanation:
2.9+2/9
A)2.8
b1) 4.5/2.8
b2)impossible because if you want to do trigonometry you can't have that is an angle (Which is 148.1) larger than 90 degrees.
but if its angle BAF then the angle would be 2.8/5.3
c)41.32
Answer:
da first one
Step-by-step explanation:
