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 units
Step-by-step explanation:
Answer:
Yes, the center of a circle is always and always equidistant from all the points on circumference because that is because what a circle is called. The distance from the center of the circle to the circumference is called the radius and it is constant for any particular circle.
Answer:
No its independent
Step-by-step explanation:
because you'll have to divide and stuff
Answer: how many square feet are there? multiple each square foot by 2
Step-by-step explanation:
