Good day ^-^ ///////////////
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:
The third option (foot/1 hour)
Step-by-step explanation:
It’s the third one.. I think
Heya !
Given expression -
Subtracting 25 both sides ,
Dividing by 3 on both sides ,
Therefore ,
